Angles and Their Measures
In trigonometry, an angle is formed by the rotation of a ray about its endpoint. The initial side of the angle is the starting position of the ray, and the terminal side is where the ray stops after rotating. Angles can be measured in different units: degrees and radians.
Units of Angle Measurement
1. Degrees: The most common unit for measuring angles. A full rotation is divided into 360 equal parts, and each part is called a degree (°).
- 1 degree = 1/360 of a full circle.
2. Radians: The SI unit for measuring angles, where one radian is the angle formed when the radius of the circle is wrapped around the circle's edge.
- 1 radian ≈ 57.2958 degrees.
- Full circle in radians = \(2\pi\) radians.
The relation between degrees and radians is:
\(180^\circ = \pi \text{ radians}\)
For conversion:
- To convert degrees to radians: \(\text{radians} = \frac{\text{degrees} \cdot \pi}{180}\)
- To convert radians to degrees: \(\text{degrees} = \frac{\text{radians} \cdot 180}{\pi}\)
Types of Angles
1. Acute Angle: An angle less than 90°.
2. Right Angle: An angle exactly equal to 90°.
3. Obtuse Angle: An angle between 90° and 180°.
4. Straight Angle: An angle exactly equal to 180°.
5. Reflex Angle: An angle between 180° and 360°.
Coterminal Angles
Coterminal angles are angles that share the same initial and terminal sides but differ by multiples of a full rotation (360° or \(2\pi\) radians). For example:
\(\text{For an angle of } 30^\circ, \text{ its coterminal angles are } 30^\circ \pm 360^\circ n \text{ (where n is an integer)}.\)
Complementary and Supplementary Angles
1. Complementary Angles: Two angles whose sum is 90°.
- Example: If \(\theta = 40^\circ\), its complementary angle is \(50^\circ\), because \(40^\circ + 50^\circ = 90^\circ\).
2. Supplementary Angles: Two angles whose sum is 180°.
- Example: If \(\theta = 120^\circ\), its supplementary angle is \(60^\circ\), because \(120^\circ + 60^\circ = 180^\circ\).
Example Problem
Q1: Convert 60° into radians.
- Solution:
\(\text{Radians} = \frac{60^\circ \cdot \pi}{180^\circ} = \frac{\pi}{3} \text{ radians}\)
Q2: Find the supplementary angle of \(110^\circ\).
- Solution: Supplementary angle = \(180^\circ - 110^\circ = 70^\circ\).
Application in IIT JEE
Angles and their measures are fundamental in solving trigonometric equations, analyzing waveforms, and modeling circular motion. Understanding radian measures is crucial for working with calculus and in physics, particularly in rotational dynamics.
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Q 1. If the arcs of the same lengths in two circles subtend angles \(65^{\circ}\) and \(110^{\circ}\) at the centre, the ratio of their radii is; |
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(d) \(21: 13\); |
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(c) \(22: 13\); |
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(b) \(22: 31\); |
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(a) \(12: 13\); |
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Q 2. A tower stands at the centre of a circular park \(A\) and \(B\) are two points on the boundary of the park such that \(A B(=a)\) subtends an angle of \(60^{\circ}\) at the foot of the tower, and the angle of elevation of the top of the tower from \(A\) or \(B\) is \(30^{\circ}\) The height of the tower is; |
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(d) \(a \sqrt{3}\); |
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(c) \(\frac{a}{\sqrt{3}}\); |
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(b) \(2 a \sqrt{3}\); |
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(a) \(\frac{2 a}{\sqrt{3}}\); |
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Q 3. A particle just clears a wall of height \(b\) at distance \(a\) and strikes the ground at a distance \(c\) from the point of projection The angle of projection is; |
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(d) \(\tan ^{-1}\left(\frac{b c}{a}\right)\); |
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(c) \(\tan ^{-1}\left(\frac{b c}{a(c-a)}\right)\); |
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(b) \(45^{\circ}\); |
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(a) \(\tan ^{-1}\left(\frac{b}{a c}\right)\); |
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Q 4. \(A B\) is a vertical pole with \(B\) at the ground level and \(A\) at the top A man finds that the angle of elevation of point \(A\) from a certain point \(C\) on the ground is \(60^{\circ}\) He moves away from the pole along the line \(B C\) to a point \(D\) such that \(C D=7 \mathrm{~m}\) From \(D\) the angle of elevation of the point \(A\) is \(45^{\circ}\) Then the height of the pole is; |
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(d) \(\frac{7 \sqrt{3}}{2} \cdot\left(\frac{1}{\sqrt{3}+1}\right) \mathrm{m}\); |
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(c) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}-1) \mathrm{m}\); |
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(b) \(\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1) \mathrm{m}\); |
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(a) \(\frac{7 \sqrt{3}}{2} \cdot\left(\frac{1}{\sqrt{3}-1}\right) \mathrm{m}\); |
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Q 5. Two stones are projected from the top of a cliff \(h\) meters high, with the same speed \(u\) so as to hit the ground at the same spot If one of the stones is projected horizontally and the other is projected at an angle \(\theta\) to the horizontal then \(\tan \theta\) equals; |
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(d) \(u \sqrt{\frac{2}{g h}}\); |
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(c) \(2 h \sqrt{\frac{u}{g}}\); |
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(b) \(2 g \sqrt{\frac{u}{h}}\); |
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(a) \(\sqrt{\frac{2 u}{g h}}\); |
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Q 6. The upper \(\left(\frac{3}{4}\right)\) th portion of a vertical pole subtends an angle \(\tan ^{-1}\left(\frac{3}{5}\right)\) at a point in the horizontal plane through its foot and at a distance 40 m from the foot A possible height of the vertical pole is; |
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(a) 20 m; |
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(b) 40 m; |
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(c) 60 m; |
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(d) 80 m; |
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Q 7. A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is \(60^{\circ}\) and when he retires 40 metre away from the tree the angle of elevation becomes \(30^{\circ}\) The breadth of the river is; |
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(a) 20 m; |
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(b) 30 m; |
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(c) 40 m; |
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(d) 60 m; |
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Q 8. The angle of elevation of the top of a vertical tower from a point \(A\), due east of it is \(45^{\circ}\) The angle of elevation of the top of the same tower from a point \(B\), due south of \(A\) is \(30^{\circ}\) If the distance between \(A\) and \(B\) is \(54 \sqrt{2} \mathrm{~m}\), then the height of the tower (in metres), is; |
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(a) 108; |
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(b) \(36 \sqrt{3}\); |
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(c) \(54 \sqrt{3}\); |
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(d) 54; |
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Q 9. \(P Q R\) is a triangular park with \(P Q=P R=200 \mathrm{~m}\) A TV tower stands at the mid-point of \(Q R\) If the angles of elevation of the top of the tower at \(P, Q\) and \(R\) are, respectively, \(45^{\circ}, 30^{\circ}\) and \(30^{\circ}\), then the height of the tower (in m ) is; |
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(a) 50; |
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(b) \(100 \sqrt{3}\); |
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(c) \(50 \sqrt{3}\); |
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(d) 100; |
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Q 10. A man is walking towards a vertical pillar in a straight path, at a uniform speed At a certain point \(A\) on the path, he observes that the angle of elevation of the top of the pillar is \(30^{\circ}\) After walking for 10 min from \(A\) in the same direction, at a point \(B\), he observes that the angle of elevation of the top of the pillar is \(60^{\circ}\) Then, the time taken (in minutes) by him, from \(B\) to reach the pillar, is; |
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(d) 20; |
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(c) 10; |
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(b) 6; |
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(a) 5; |
Symmetry of Trigonometric Functions
The symmetry properties of trigonometric functions are essential when solving equations or simplifying expressions.
- Symmetry with respect to the origin (odd functions):
\(\sin(-\theta) = -\sin(\theta)\)
\(\tan(-\theta) = -\tan(\theta)\)
These functions exhibit symmetry with respect to the origin, which means they are symmetric about the point \((0, 0)\).
- Symmetry with respect to the y-axis (even functions):
\(\cos(-\theta) = \cos(\theta)\)
\(\sec(-\theta) = \sec(\theta)\)
These functions are symmetric with respect to the y-axis, meaning the graph of these functions mirrors itself on either side of the y-axis.
4. Range of Trigonometric Functions
The range is the set of values a function can take.
- Sine and Cosine:
\(-1 \leq \sin(\theta) \leq 1 \)
\(-1 \leq \cos(\theta) \leq 1 \)
- Tangent and Cotangent:
\(\text{Range of } \tan(\theta): (-\infty, +\infty)\)
\(\text{Range of } \cot(\theta): (-\infty, +\infty)\)
- Secant and Cosecant:
\(\text{Range of } \sec(\theta): (-\infty, -1) \cup (1, \infty)\)
\(\text{Range of } \csc(\theta): (-\infty, -1) \cup (1, \infty)\)
5. Domain of Trigonometric Functions
The domain is the set of input values (angles) for which a function is defined.
- Sine and Cosine: The domain is all real numbers:
\(\text{Domain of } \sin(\theta) \text{ and } \cos(\theta): (-\infty, +\infty)\)
- Tangent and Secant: Tangent is undefined where \(\cos(\theta) = 0\)\), which occurs at:
\(\theta = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}\)
- Cotangent and Cosecant: Cotangent is undefined where \(\sin(\theta) = 0\), which occurs at:
\(\theta = n\pi, n \in \mathbb{Z}\)
6. Fundamental Identities
- Pythagorean Identities:
\( \sin^2(\theta) + \cos^2(\theta) = 1 \)
\( 1 + \tan^2(\theta) = \sec^2(\theta)\)
\( 1 + \cot^2(\theta) = \csc^2(\theta) \)
Example Problem
Q1: Prove that \( \sin^2(\theta) + \cos^2(\theta) = 1 \) for \(\theta = 30^\circ\).
- Solution:
\(\sin(30^\circ) = \frac{1}{2}, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2}\)
Substituting these values into the identity:
\(\left(\frac{1}{2}\right)^2 +\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1\)
Hence, the identity holds true.
Application in IIT JEE
The properties of trigonometric functions are crucial in solving complex problems related to waveforms, oscillations, circular motion, and in simplifying expressions in calculus. These properties are also extensively tested in IIT JEE problems, particularly in the trigonometry and calculus sections.
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Q 1. Q8. A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length \(x\). The maximum area enclosed by the park is:; |
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(A) \(\frac{3}{2} x^{2}\); |
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(B) \(\sqrt{\frac{x^{3}}{8}}\); |
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(C) \(\pi x^{2}\); |
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(D) \(\frac{1}{2} x^{2}\); |
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Q 2. Q14. Evaluate \(\cot \left(\sum_{n=1}^{19} \cot ^{-1}\left(1+\sum_{p=1}^{n} 2 p\right)\right)\); |
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(D) \(\frac{22}{23}\); |
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(C) \(\frac{23}{22}\); |
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(B) \(\frac{19}{21}\); |
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(A) \(\frac{21}{19}\); |
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Q 3. If the reflection of the ellipse \(\frac{(x-4)^{2}}{16}+\frac{(y-3)^{2}}{9}=1\) in the line mirror \(x-y-2=0\) is \(k_{1} x^{2}+k_{2} y^{2}-160 x-36 y+292=0\), then \(k_{1}+k_{2}\) is equal to; |
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(D) 105; |
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(C) 45; |
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(B) 25; |
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(A) 22; |
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Q 4. The value of \(\lambda\), for which the line \(2 x-\frac{8}{3} \lambda y=-3\) is a normal to the conic \(x^{2}+\frac{y^{2}}{4}=1\) is:; |
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(D) \(\frac{\sqrt{3}}{2}\); |
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(C) \(-\frac{\sqrt{3}}{4}\); |
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(B) \(\frac{1}{2}\); |
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(A) \(\frac{3}{8}\); |
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Q 5. Let \(L\) be a common tangent line to the curves \(4 x^{2}+9 y^{2}=36\) and \((2 x)^{2}+(2 y)^{2}=31\). Then the square of the slope of the line \(L\) is; |
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(D) 12; |
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(C) 3; |
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(B) 4; |
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(A) 2; |
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Q 6. The foci of the basic terms of conics \(25 x^{2}+16 y^{2}-150 x=175\) are; |
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(D) \((0, \pm 1)\); |
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(C) \((3, \pm 3)\); |
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(B) \((0, \pm 2)\); |
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(A) \((0, \pm 3)\); |
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Q 7. Q2. A bird is perched on the top of a tree \(20 \mathrm{~m}\) high and its elevation from a point on the ground is \(45^{\circ}\). It flies off horizontally straight away from the observer and in one second the elevation of the bird is reduced to \(30^{\circ}\). The speed of the bird is -; |
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(D) None of these; |
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(C) \(12 \mathrm{~m} / \mathrm{s}\); |
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(B) \(17.71 \mathrm{~m} / \mathrm{s}\); |
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(A) \(14.64 \mathrm{~m} / \mathrm{s}\); |
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Q 8. The foci of the basic terms of conics \(25 x^{2}+16 y^{2}-150 x=175\) are; |
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(A) \((0, \pm 3)\); |
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(B) \((0, \pm 2)\); |
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(C) \((3, \pm 3)\); |
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(D) \((0, \pm 1)\); |
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Q 9. Q14. Evaluate \(\cot \left(\sum_{n=1}^{19} \cot ^{-1}\left(1+\sum_{p=1}^{n} 2 p\right)\right)\); |
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(A) \(\frac{21}{19}\); |
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(B) \(\frac{19}{21}\); |
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(C) \(\frac{23}{22}\); |
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(D) \(\frac{22}{23}\); |
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Q 10. Q8. A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length \(x\). The maximum area enclosed by the park is:; |
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(A) \(\frac{3}{2} x^{2}\); |
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(B) \(\sqrt{\frac{x^{3}}{8}}\); |
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(C) \(\pi x^{2}\); |
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(D) \(\frac{1}{2} x^{2}\); |
Solution of Triangles
The solution of triangles is an important topic in trigonometry, particularly in solving real-world problems involving angles and distances. This concept primarily deals with determining the unknown sides or angles of a triangle, using known sides and angles. The methods to solve triangles depend on whether the triangle is right-angled or oblique (non-right-angled).
Types of Triangles
1. Right-Angled Triangle: A triangle with one angle equal to \(90^\circ\).
2. Oblique Triangle: A triangle that does not have a \(90^\circ\) angle (i.e., all angles are acute or obtuse).
1. Solution of Right-Angled Triangles
For a right-angled triangle, we use the basic trigonometric ratios to solve for unknown sides or angles. Let’s consider a right triangle where one angle is \(90^\circ\), and the sides are labeled as follows:
- Hypotenuse: The side opposite the right angle (longest side).
- Adjacent: The side adjacent to the angle under consideration.
- Opposite: The side opposite the angle under consideration.
The trigonometric ratios are:
- Sine: \(\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}\)
- Cosine: \(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
- Tangent: \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\)
Using these ratios, the unknown sides or angles can be calculated if one side and one non-right angle are known.
Example: Solving a Right Triangle
Given: A right-angled triangle with one angle \( \theta = 30^\circ \), and the hypotenuse is 10 units.
- To find the opposite side:
\(\sin(30^\circ) = \frac{\text{Opposite}}{10} \)
\( \frac{1}{2} = \frac{\text{Opposite}}{10} \quad \Rightarrow \quad \text{Opposite} = 5 \text{ units} \)
- To find the adjacent side:
\(\cos(30^\circ) = \frac{\text{Adjacent}}{10}\)
\(\frac{\sqrt{3}}{2} = \frac{\text{Adjacent}}{10} \quad \Rightarrow \quad \text{Adjacent} = 5\sqrt{3} \text{ units}\)
2. Solution of Oblique Triangles
For oblique triangles, we use two primary rules:
1. Law of Sines:
\(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)
where \(a\), \(b\), and \(c\) are the sides of the triangle, and \(A\), \(B\), and \(C\) are the angles opposite these sides. This law is used when two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA) are known.
2. Law of Cosines:
\(c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \)
where \(c\) is the side opposite angle \(C\). The Law of Cosines is used when two sides and the included angle (SAS) or all three sides (SSS) are known.
Example: Law of Sines
Given: \(A = 30^\circ\), \(B = 45^\circ\), and \(a = 5 \text{ units}\). Find side \(b\).
- Using the Law of Sines:
\(\frac{a}{\sin A} = \frac{b}{\sin B}\)
\(\frac{5}{\sin(30^\circ)} = \frac{b}{\sin(45^\circ)} \)
\(\frac{5}{\frac{1}{2}} = \frac{b}{\frac{\sqrt{2}}{2}} \quad \Rightarrow \quad 10 = \frac{b\sqrt{2}}{2}\)
\( b = \frac{10 \times \sqrt{2}}{2} = 5\sqrt{2} \text{ units}\)
Example: Law of Cosines
Given: \(a = 7\), \(b = 8\), and \(C = 60^\circ\). Find side \(c\).
- Using the Law of Cosines:
\( c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \)
\( c^2 = 7^2 + 8^2 - 2 \times 7 \times 8 \times \cos(60^\circ) \)
\( c^2 = 49 + 64 - 56 \)
\( c^2 = 57 \quad \Rightarrow \quad c = \sqrt{57} \approx 7.55 \text{ units} \)
3. Area of a Triangle
For any triangle, the area can be calculated using various methods, depending on the known quantities.
1. Using Base and Height:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
2. Using the Law of Sines (when two sides and the included angle are known):
\(\text{Area} = \frac{1}{2} \times a \times b \times \sin C\)
Example Problem: Area of a Triangle
Given: \(a = 7\), \(b = 8\), and \(C = 45^\circ\). Find the area.
- Solution:
\(\text{Area} = \frac{1}{2} \times 7 \times 8 \times \sin(45^\circ)\)
\(\text{Area} = \frac{1}{2} \times 7 \times 8 \times \frac{\sqrt{2}}{2} = 19.79 \text{ square units}\)
Application in IIT JEE
The solution of triangles is highly relevant in IIT JEE, especially in questions involving geometry, physics (especially mechanics and wave motion), and calculus. Many problems in coordinate geometry and trigonometry require the use of these laws and properties to determine unknown quantities.
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Q 1. \(A B C\) is a triangle Forces \(\vec{P}, \vec{Q}, \vec{R}\) acting along \(I A, I B\) and \(I C\) respectively are in equilibrium, where \(I\) is the incentre of \(\triangle A B C\) Then \(P: Q: R\) is; |
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(a) \(\sin A: \sin B: \sin C\); |
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(b) \(\sin \frac{A}{2}: \sin \frac{B}{2}: \sin \frac{C}{2}\); |
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(c) \(\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2}\); |
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(d) \(\cos A: \cos B: \cos C\); |
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Q 2. In a \(\triangle A B C, \frac{a}{b}=2+\sqrt{3}\) and \(\angle C=60^{\circ}\) The ordered pair \((\angle A, \angle B)\) is equal to; |
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(a) \(\left(15^{\circ}, 105^{\circ}\right)\); |
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(b) \(\left(105^{\circ}, 15^{\circ}\right)\); |
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(c) \(\left(45^{\circ}, 75^{\circ}\right)\); |
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(d) \(\left(75^{\circ}, 45^{\circ}\right)\); |
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Q 3. A triangle with vertices \((4,0),(-1,-1),(3,5)\) is; |
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(b) isosceles but not right angled.; |
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(c) right angled but not isosceles.; |
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(d) neither right angled nor isosceles.; |
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(a) isosceles and right angled.; |
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Q 4. In a triangle \(P Q R, \angle R=\frac{\pi}{2}\) If \(\tan \left(\frac{P}{2}\right)\) and \(\tan \left(\frac{Q}{2}\right)\) are the roots of \(a x^{2}+b x+c=0, a \neq 0\) then; |
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(a) \(a=b+c\); |
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(b) \(c=a+b\); |
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(c) \(b=c\); |
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(d) \(b=a+c\); |
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Q 5. In a triangle \(A B C\), medians \(A D\) and \(B E\) are drawn If \(A D=4, \angle D A B=\frac{\pi}{6}\) and \(\angle A B E=\frac{\pi}{3}\), then the area of the \(\triangle A B C\) is; |
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(a) \(\frac{8}{3}\); |
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(b) \(\frac{16}{3}\); |
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(c) \(\frac{32}{3}\); |
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(d) \(\frac{64}{3}\); |
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Q 6. The incentre of the triangle with vertices \((1, \sqrt{3}),(0,0)\) and \((2,0)\) is; |
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(a) \(\left(1, \frac{\sqrt{3}}{2}\right)\); |
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(b) \(\left(\frac{2}{3}, \frac{1}{\sqrt{3}}\right)\); |
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(c) \(\left(\frac{2}{3}, \frac{\sqrt{3}}{2}\right)\); |
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(d) \(\left(1, \frac{1}{\sqrt{3}}\right)\); |
Trigonometric Equations
Trigonometric equations are equations that involve trigonometric functions, such as sine, cosine, tangent, etc. Solving these equations requires using properties and identities of trigonometric functions, as well as their periodic nature.
1. General Form of Trigonometric Equations
A basic trigonometric equation can take the form:
\(a \cdot \sin(x) + b \cdot \cos(x) = c\)
or more complex forms involving multiple trigonometric terms. The goal is to find the value(s) of the unknown variable \(x\) (usually an angle) that satisfies the equation.
2. Periodicity of Trigonometric Functions in Equations
Since trigonometric functions are periodic, the solutions to trigonometric equations are often periodic as well. For example:
- The sine and cosine functions repeat every \(2\pi\) radians.
- The tangent and cotangent functions repeat every \(\pi\) radians.
Therefore, if \(x_0\) is a solution to a trigonometric equation involving sine or cosine, then:
\(x = x_0 + 2n\pi, \quad n \in \mathbb{Z}\)
For tangent or cotangent:
\(x = x_0 + n\pi, \quad n \in \mathbb{Z}\)
3. Methods of Solving Trigonometric Equations
There are several techniques used to solve trigonometric equations:
a. Solving Simple Equations
For basic equations such as \(\sin(x) = k\) or \(\cos(x) = k\), solutions can be found by using the inverse trigonometric functions and the periodicity of the functions.
- Example: Solve \(\sin(x) = \frac{1}{2}\) for \(0 \leq x \leq 2\pi\).
- From known values, \(\sin(x) = \frac{1}{2}\) at \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).
- Thus, the solutions are:
\(x = \frac{\pi}{6}, \frac{5\pi}{6}\)
Since sine is periodic, other solutions can be written as:
\( x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi, \quad n \in \mathbb{Z}\)
b. Using Trigonometric Identities
Sometimes, trigonometric equations require using identities such as the Pythagorean identities, double angle identities, or sum-to-product formulas to simplify or transform the equation.
- Example: Solve \(\sin(2x) = \cos(x)\).
- First, use the identity \(\sin(2x) = 2\sin(x)\cos(x)\):
\(2\sin(x)\cos(x) = \cos(x)\)
- If \(\cos(x) \neq 0\), divide both sides by \(\cos(x)\):
\(2\sin(x) = 1 \quad \Rightarrow \quad \sin(x) = \frac{1}{2}\)
- From known values, \(\sin(x) = \frac{1}{2}\) at \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).
- Thus, the solutions are:
\(x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi, \quad n \in \mathbb{Z}\)
- Additionally, check for \(\cos(x) = 0\), which occurs at \(x = \frac{\pi}{2} + n\pi\).
c. Quadratic Trigonometric Equations
Quadratic equations involving trigonometric functions can be solved using techniques similar to solving quadratic equations in algebra.
- Example: Solve \(2\sin^2(x) - 3\sin(x) + 1 = 0\).
- Let \(\sin(x) = t\). The equation becomes:
\(2t^2 - 3t + 1 = 0\)
- Solve the quadratic equation using the quadratic formula:
\(t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}\)
- So, \(t = 1\) or \(t = \frac{1}{2}\).
- Now, solve for \(x\):
- If \(\sin(x) = 1\), then \(x = \frac{\pi}{2} + 2n\pi\).
- If \(\sin(x) = \frac{1}{2}\), then \(x = \frac{\pi}{6} + 2n\pi\) or \(x = \frac{5\pi}{6} + 2n\pi\).
4. Special Angles and Their Values
It is important to memorize the values of sine, cosine, and tangent for special angles such as \(0^\circ\), \(30^\circ\), \(45^\circ\), \(60^\circ\), and \(90^\circ\), as they frequently appear in trigonometric equations.
- Example Values:
\(\sin(0^\circ) = 0, \quad \sin(30^\circ) = \frac{1}{2}, \quad \sin(45^\circ) = \frac{\sqrt{2}}{2}, \quad \sin(60^\circ) = \frac{\sqrt{3}}{2}, \quad \sin(90^\circ) = 1\)
\(\cos(0^\circ) = 1, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \cos(45^\circ) = \frac{\sqrt{2}}{2}, \quad \cos(60^\circ) = \frac{1}{2}, \quad \cos(90^\circ) = 0\)
5. Common Mistakes to Avoid
- Not considering the periodic nature of the solutions. Always add the periodic term \(2n\pi\) for sine and cosine, and \(n\pi\) for tangent and cotangent.
- Forgetting the range of trigonometric functions. For example, \(\sin(x)\) and \(\cos(x)\) can only take values between \(-1\) and \(1\).
- Neglecting multiple solutions. For instance, \(\sin(x) = \frac{1}{2}\) has solutions at both \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\).
Example Problem
Q1: Solve \( \cos(2x) = \frac{1}{2} \) for \(0 \leq x \leq 2\pi\).
- Solution:
- From known values, \(\cos(2x) = \frac{1}{2}\) when \(2x = \frac{\pi}{3}\) or \(2x = \frac{5\pi}{3}\).
- Therefore, \(x = \frac{\pi}{6}\) or \(x = \frac{5\pi}{6}\).
- Since cosine is periodic, other solutions are:
\(x = \frac{\pi}{6} + n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + n\pi, \quad n \in \mathbb{Z}\)
Application in IIT JEE
Trigonometric equations frequently appear in the IIT JEE Mains and Advanced exams. They often involve simplifying complex expressions or solving multi-step problems using identities and inverse trigonometric functions.
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Q 1. If \(\tan \theta=-\frac{1}{\sqrt{3}}\), then general solution of the equation is; |
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(d) \(\mathrm{n} \pi-\frac{\pi}{6}, \mathrm{n} \in \mathrm{I}\); |
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(c) \(2 \mathrm{n} \pi-\frac{\pi}{6}, \mathrm{n} \in \mathrm{I}\); |
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(b) \(\mathrm{n} \pi+\frac{\pi}{6}, \mathrm{n} \in \mathrm{I}\); |
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(a) \(2 \mathrm{n} \pi+\frac{\pi}{6}, \mathrm{n} \in \mathrm{I}\); |
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Q 2. The general value of \(\theta\) satisfying the equation \(\tan \theta+\tan \left(\frac{\pi}{2}-\theta\right)=2\), is; |
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(d) \(\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}\); |
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(c) \(2 \mathrm{n} \pi \pm \frac{\pi}{4}\); |
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(b) \(\mathrm{n} \pi+\frac{\pi}{4}\); |
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(a) \(\mathrm{n} \pi \pm \frac{\pi}{4}\); |
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Q 3. General solution of the equation \(\tan \theta \tan 2 \theta=1\) is given by; |
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(d) \(\mathrm{n} \pi \pm \frac{\pi}{6}, \mathrm{n} \in \mathrm{I}\); |
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(c) \(\mathrm{n} \pi-\frac{\pi}{6}, \mathrm{n} \in \mathrm{I}\); |
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(b) \(\mathrm{n} \pi+\frac{\pi}{6}, \mathrm{n} \in \mathrm{I}\); |
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(a) \((2 \mathrm{n}+1) \frac{\pi}{4}, \mathrm{n} \in \mathrm{I}\); |
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Q 4. Solution of the equation \(3 \tan (\theta-15)=\tan (\theta+15)\) is; |
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(d) \(\theta=\mathrm{n} \pi-\frac{\pi}{4}\); |
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(c) \(\theta=\mathrm{n} \pi-\frac{\pi}{3}\); |
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(b) \(\theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{3}\); |
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(a) \(\theta=\frac{\mathrm{n} \pi}{2}+(-1)^{\mathrm{n}} \frac{\pi}{4}\); |
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Q 5. Assertion : The solution of the equation \(\tan \theta+\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)=3\) is \(\theta=\frac{\mathrm{n} \pi}{3}+\frac{\pi}{12}, \mathrm{n} \in \mathrm{I}\). Reason : If \(\tan \theta=\tan \alpha\), then \(\theta=\mathrm{n} \pi+\alpha, \mathrm{n} \in \mathrm{I}\).; |
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(d) Assertion is incorrect, reason is correct.; |
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(c) Assertion is correct, reason is incorrect; |
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(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion; |
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(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.; |
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Q 6. The number of solutions of the given equation \(\tan \theta+\sec \theta=\sqrt{3}\), where \(0 \leq \theta \leq 2 \pi\) is; |
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(a) 0; |
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(b) 1; |
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(c) 2; |
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(d) 3; |
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Q 1. In a triangle \(A B C, a=4, b=3, \angle A=60^{\circ}\), then \(c\) is the root of the equation; |
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(a) \(c^{2}-3 c-7=0\); |
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(b) \(c^{2}+3 c+7=0\); |
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(c) \(c^{2}-3 c+7=0\); |
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(d) \(c^{2}+3 c-7=0\); |
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for every value of the variables involved. These identities are essential for simplifying complex trigonometric expressions, solving equations, and analyzing functions in calculus and physics. There are several key categories of identities, each with its own applications and use cases.
1. Fundamental Trigonometric Identities
These are the most basic identities, which form the foundation of trigonometry.
- Reciprocal Identities:
\(\sin(x) = \frac{1}{\csc(x)}, \quad \cos(x) = \frac{1}{\sec(x)}, \quad \tan(x) = \frac{1}{\cot(x)}\)
\(\csc(x) = \frac{1}{\sin(x)}, \quad \sec(x) = \frac{1}{\cos(x)}, \quad \cot(x) = \frac{1}{\tan(x)}\)
- Quotient Identities:
\(\tan(x) = \frac{\sin(x)}{\cos(x)}, \quad \cot(x) = \frac{\cos(x)}{\sin(x)}\)
2. Pythagorean Identities
The Pythagorean identities are derived from the Pythagorean theorem and apply to any angle \(x\).
- Primary Pythagorean Identity:
\(\sin^2(x) + \cos^2(x) = 1\)
This identity is used extensively in simplifying trigonometric expressions and in solving equations.
- Other Pythagorean Identities (derived from the primary identity):
\(1 + \tan^2(x) = \sec^2(x)\)
\(1 + \cot^2(x) = \csc^2(x)\)
3. Co-Function Identities
The co-function identities relate the values of trigonometric functions at complementary angles (i.e., angles that add up to \(90^\circ\) or \(\frac{\pi}{2}\) radians).
- Co-Function Identities:
\(\sin\left(\frac{\pi}{2} - x\right) = \cos(x), \quad \cos\left(\frac{\pi}{2} - x\right) = \sin(x)\)
\(\tan\left(\frac{\pi}{2} - x\right) = \cot(x), \quad \cot\left(\frac{\pi}{2} - x\right) = \tan(x)\)
\(\sec\left(\frac{\pi}{2} - x\right) = \csc(x), \quad \csc\left(\frac{\pi}{2} - x\right) = \sec(x)\)
4. Even-Odd Identities
These identities reflect the symmetry properties of trigonometric functions, based on whether the functions are even or odd.
- Even Functions: Cosine and secant are even functions, meaning:
\(\cos(-x) = \cos(x), \quad \sec(-x) = \sec(x)\)
- Odd Functions: Sine, tangent, cotangent, and cosecant are odd functions, meaning:
\(\sin(-x) = -\sin(x), \quad \tan(-x) = -\tan(x), \quad \csc(-x) = -\csc(x), \quad \cot(-x) = -\cot(x)\)
5. Sum and Difference Identities
These identities are used to find the sine, cosine, and tangent of sums or differences of two angles. They are crucial in solving trigonometric equations and simplifying expressions.
- Sine of Sum and Difference:
\(\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)\)
\(\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)\)
- Cosine of Sum and Difference:
\(\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)\)
\(\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)\)
- Tangent of Sum and Difference:
\(\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}\)
\(\tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}\)
6. Double Angle Identities
Double angle identities are used when an expression involves the sine, cosine, or tangent of twice an angle.
- Double Angle Formulas:
\(\sin(2x) = 2\sin(x)\cos(x)\)
\(\cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x)\)
\(\tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)}\)
7. Half Angle Identities
These identities are useful for finding the trigonometric functions of half-angles.
- Sine, Cosine, and Tangent of Half-Angles:
\(\sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos(x)}{2}}\)
\(\cos\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 + \cos(x)}{2}}\)
\(\tan\left(\frac{x}{2}\right) = \frac{\sin(x)}{1 + \cos(x)} = \frac{1 - \cos(x)}{\sin(x)}\)
8. Product-to-Sum and Sum-to-Product Identities
These identities are used to convert products of trigonometric functions into sums or differences and vice versa. They are often applied when simplifying integrals or solving trigonometric equations.
- Product-to-Sum Formulas:
\(\sin(A)\sin(B) = \frac{1}{2}(\cos(A - B) - \cos(A + B))\)
\(\cos(A)\cos(B) = \frac{1}{2}(\cos(A - B) + \cos(A + B))\)
\(\sin(A)\cos(B) = \frac{1}{2}(\sin(A + B) + \sin(A - B))\)
- Sum-to-Product Formulas:
\(\sin(A) + \sin(B) = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)\)
\(\sin(A) - \sin(B) = 2\cos\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right)\)
\(\cos(A) + \cos(B) = 2\cos\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)\)
Example Problem
Q1: Simplify \(\sin(75^\circ)\cos(15^\circ) + \cos(75^\circ)\sin(15^\circ)\).
- Solution:
Use the sum identity for sine:
\(\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)\)
Therefore:
\(\sin(75^\circ)\cos(15^\circ) + \cos(75^\circ)\sin(15^\circ) = \sin(75^\circ + 15^\circ) = \sin(90^\circ) = 1\)
Application in IIT JEE
Trigonometric identities are heavily tested in both IIT JEE Mains and Advanced. They are used to simplify complex expressions, solve equations, and prove results. Mastery of these identities allows students to tackle a wide range of problems in algebra, calculus, and even physics.
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Q 1. Q3. \(\cos 40^{\circ} \cos 80^{\circ} \cos 160^{\circ}=\); |
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(A) \(\frac{-1}{2}\); |
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(B) \(\frac{-1}{4}\); |
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(C) \(\frac{-1}{8}\); |
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(D) \(\frac{-1}{16}\); |
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Q 2. Q2. If \(\cos h(2 x)=199\), then \(\cot h x=\); |
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(A) \(\frac{5}{3 \sqrt{11}}\); |
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(B) \(\frac{5}{6 \sqrt{11}}\); |
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(C) \(\frac{7}{3 \sqrt{11}}\); |
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(D) \(\frac{10}{3 \sqrt{11}}\); |
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Q 3. Q1. If \(\theta\) lies in the second quadrant and \(3 \tan \theta+4=0\), then the value of \(2 \cot \theta-5 \cos \theta+\sin \theta\) is \(\frac{6}{k}\) then find \(k\).; |
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(A) \(\frac{23}{10}\); |
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(B) \(\frac{24}{10}\); |
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(C) \(\frac{23}{11}\); |
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(D) \(\frac{23}{9}\); |
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Q 4. Q6. The value of \(\cos 65^{\circ} \cos 55^{\circ} \cos 5^{\circ}=\); |
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(A) \(\frac{\sqrt{3}-1}{4 \sqrt{2}}\); |
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(B) \(\frac{\sqrt{3}-1}{8 \sqrt{2}}\); |
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(C) \(\frac{\sqrt{3}+1}{4 \sqrt{2}}\); |
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(D) \(\frac{\sqrt{3}+1}{8 \sqrt{2}}\); |
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Q 5. Q9. The value of\n\(\cos \left(-89^{\circ}\right)+\cos \left(-87^{\circ}\right)+\cos (-85)^{\circ}+\cdots+\cos \left(85^{\circ}\right)+\cos \left(87^{\circ}\right)+\cos \left(89^{\circ}\right)=\); |
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(A) \(\operatorname{cosec} 1^{\circ}\); |
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(B) \(\sec 1^{\circ}\); |
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(C) \(2 \sec 1^{\circ}\); |
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(D) \(2 \operatorname{cosec} 1^{\circ}\); |
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Q 6. Q10. The value of \(\cos \frac{\pi}{65} \cos \frac{2 \pi}{65} \cos \frac{4 \pi}{65} \ldots \cos \frac{32 \pi}{65}\) is; |
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(A) \(\frac{1}{32}\); |
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(B) \(\frac{1}{64}\); |
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(C) \(-\frac{1}{32}\); |
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(D) \(-\frac{1}{64}\); |
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Q 7. Q13. If \(\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha\), then \(4 x^{2}-4 x y \cos \alpha+y^{2}\) is equal to; |
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(A) 4; |
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(B) \(2 \sin 2 \alpha \); |
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(C) \(-4 \sin ^{2} \alpha\); |
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(D) \(4 \sin ^{2} \alpha\); |
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Q 8. Q12. \(2 \pi-\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65}\right)\) is equal to :; |
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(C) \(\frac{3 \pi}{2}\); |
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(D) \(\frac{7 \pi}{4}\); |
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(B) \(\frac{5 \pi}{4}\); |
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(A) \(\frac{\pi}{2}\); |
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Q 9. Q7. If \(3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1} \frac{1-x^{2}}{1+x^{2}}+2 \tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3}\), then the value of \(x\) lying in \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) is; |
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(D) \(-\frac{\sqrt{3}}{2}\); |
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(C) \(\frac{\sqrt{3}}{2}\); |
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(B) \(\frac{1}{\sqrt{3}}\); |
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(A) \(\frac{1}{2}\); |
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Q 10. Q4. If \(\sin ^{-1} \sin 17+\cos ^{-1} \cos 27+\tan ^{-1} \tan 37\) is equal to \(k-\lambda \pi\), then value of \(k+\lambda\) is; |
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(D) 64; |
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(C) 63; |
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(B) 62; |
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(A) 61; |
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Q 1. Q4. If \(\cosh (x)=\frac{5}{4}\), then \(\cosh (3 x)=\); |
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(A) \(\frac{61}{16}\); |
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(B) \(\frac{63}{16}\); |
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(C) \(\frac{65}{16}\); |
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(D) \(\frac{61}{63}\); |
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Q 2. Q4. If \(\cosh (x)=\frac{5}{4}\), then \(\cosh (3 x)=\); |
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(A) \(\frac{61}{16}\); |
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(B) \(\frac{63}{16}\); |
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(C) \(\frac{65}{16}\); |
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(D) \(\frac{61}{63}\); |
Trigonometric Identities, Periodic Functions
Trigonometric functions, such as sine, cosine, and tangent, are periodic functions, meaning their values repeat after a certain interval. The periodicity of these functions plays a crucial role in simplifying expressions, solving equations, and analyzing patterns in mathematics and physics. In this section, we’ll explore periodic functions in detail, along with their corresponding identities and applications.
1. Definition of Periodic Functions
A function \( f(x) \) is called periodic if there exists a positive number \( T \), called the period, such that:
\(f(x + T) = f(x) \quad \text{for all } x \in \mathbb{R}\)
The smallest such positive value of \( T \) is called the fundamental period of the function.
2. Periodicity of Trigonometric Functions
The trigonometric functions exhibit the following periods:
- Sine and Cosine: Both the sine and cosine functions have a fundamental period of \( 2\pi \) radians (or 360°). This means their values repeat every \( 2\pi \) radians. Mathematically:
\(\sin(x + 2\pi) = \sin(x)\)
\(\cos(x + 2\pi) = \cos(x)\)
- Tangent and Cotangent: The tangent and cotangent functions have a fundamental period of \( \pi \) radians (or 180°). This is due to the fact that these functions repeat their values after every \( \pi \) radians. Mathematically:
\(\tan(x + \pi) = \tan(x)\)
\(\cot(x + \pi) = \cot(x)\)
- Secant and Cosecant: The secant and cosecant functions have the same period as their reciprocal functions (cosine and sine, respectively). Thus, they also have a period of \( 2\pi \) radians:
\(\sec(x + 2\pi) = \sec(x)\)
\(\csc(x + 2\pi) = \csc(x)\)
3. Periodicity of Composite Trigonometric Functions
In cases where the argument of a trigonometric function is modified (e.g., \( \sin(3x) \), \( \cos(2x) \)), the period of the function changes.
- Example 1: The function \( \sin(3x) \) has a period of \( \frac{2\pi}{3} \), because the argument \( x \) is multiplied by 3. The period of a function like \( \sin(kx) \) is given by:
\(\text{Period of } \sin(kx) = \frac{2\pi}{k}\)
- Example 2: Similarly, the function \( \cos\left(\frac{x}{2}\right) \) has a period of \( 4\pi \), because the argument is divided by 2:
\(\text{Period of } \cos\left(\frac{x}{2}\right) = 2\pi \times 2 = 4\pi\)
4. Shifted Trigonometric Functions
A shifted trigonometric function occurs when a constant is added to or subtracted from the argument of the function. This results in a horizontal shift of the function’s graph but does not affect its period.
- Example: The function \( \sin(x - \frac{\pi}{4}) \) is a sine function shifted to the right by \( \frac{\pi}{4} \) radians, but its period remains \( 2\pi \).
5. Important Trigonometric Periodic Identities
There are several important identities involving the periodicity of trigonometric functions. These identities are helpful for simplifying expressions and solving equations:
- Sine and Cosine Identities:
\(\sin(x + 2\pi) = \sin(x)\)
\(\cos(x + 2\pi) = \cos(x)\)
- Tangent and Cotangent Identities:
\(\tan(x + \pi) = \tan(x)\)
\(\cot(x + \pi) = \cot(x)\)
- Periodicity of Sum and Difference of Angles: For any integer \( n \), the following holds:
\(\sin(x + 2n\pi) = \sin(x)\)
\(\cos(x + 2n\pi) = \cos(x)\)
6. Applications of Periodic Trigonometric Functions
The periodic nature of trigonometric functions is essential in many fields, particularly in the study of waves, oscillations, and circular motion. Some key applications include:
- Harmonic Motion: The sine and cosine functions describe simple harmonic motion, which is the motion of a particle oscillating along a straight line (such as a mass on a spring or a pendulum). The position of the particle can be modeled as:
\(x(t) = A \sin(\omega t + \phi)\)
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency (related to the period by \( T = \frac{2\pi}{\omega} \)),
- \( t \) is time, and
- \( \phi \) is the phase shift.
- Waveforms: In physics, the sine and cosine functions describe periodic waves, such as sound waves or light waves. The periodicity of these functions is directly related to the wavelength of the wave.
- Fourier Series: The periodicity of trigonometric functions allows them to form the basis of Fourier series, which are used to represent periodic functions as a sum of sine and cosine terms. This is extremely useful in signal processing, where periodic signals can be broken down into their constituent frequencies.
7. General Solutions to Trigonometric Equations
Due to the periodicity of trigonometric functions, the solutions to equations involving these functions are also periodic. For example:
- Example 1: Solve \(\sin(x) = \frac{1}{2} \).
The general solutions are:
\(x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi, \quad n \in \mathbb{Z}\)
This means that the solutions repeat every \( 2\pi \) radians, as expected from the periodicity of the sine function.
- Example 2: Solve \( \cos(x) = -1 \).
The general solution is:
\(x = \pi + 2n\pi, \quad n \in \mathbb{Z}\)
Since the cosine function has a period of \( 2\pi \), the solutions occur at multiples of \( \pi \) plus a full cycle.
8. Example Problems Involving Periodic Functions
Let's work through a few problems that involve periodic trigonometric functions:
- Problem 1: Find the general solution of \( \tan(2x) = 1 \).
- Solution:
First, solve for \( 2x \):
\(2x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}\)
Divide by 2 to get the solution for \( x \):
\(x = \frac{\pi}{8} + \frac{n\pi}{2}, \quad n \in \mathbb{Z}\)
This solution represents the periodicity of the tangent function, which repeats every \( \pi \) radians.
- Problem 2: Solve \( \cos(3x) = 1 \) for all solutions.
- Solution:
First, solve for \( 3x \):
\(3x = 2n\pi, \quad n \in \mathbb{Z}\)
Divide by 3 to get the solution for \( x \):
\(x = \frac{2n\pi}{3}, \quad n \in \mathbb{Z}\)
The function \( \cos(3x) \) has a period of \( \frac{2\pi}{3} \), so the general solution repeats every \( \frac{2\pi}{3} \).
9. Key Points to Remember About Periodic Functions
- The fundamental period is the smallest value of \( T \) such that \( f(x + T) = f(x) \).
- Trigonometric functions repeat their values every \( 2\pi \) (for sine, cosine, secant, and cosecant) or \( \pi \) (for tangent and cotangent).
- The period changes when the argument of the trigonometric function is multiplied by a constant \( k \), resulting in a period of \( \frac{2\pi}{k} \) or \( \frac{\pi}{k} \).
- Shifted functions do not change their period, but the graph is displaced horizontally.
- Shifted functions do not change their period, but the graph is displaced horizontally.
- The general solution to trigonometric equations is based on the periodicity of the function. When solving these equations, you must include all possible solutions by considering the periodic nature of the function.
10. Applications in IIT JEE
The concept of periodicity is fundamental in many areas of the IIT JEE syllabus, particularly in topics related to oscillations, waves, alternating current circuits, and calculus. Trigonometric identities that exploit the periodic nature of functions are widely used in simplifying integrals, solving equations, and analyzing graphs.
Example Problem:
Q1: Find the general solution for the equation \( \sin(2x) = \sin(x) \).
- Solution:
First, use the identity for sine differences:
\(\sin(2x) - \sin(x) = 2\cos\left(\frac{3x}{2}\right)\sin\left(\frac{x}{2}\right) = 0\)
Therefore, either:
\(\cos\left(\frac{3x}{2}\right) = 0 \quad \text{or} \quad \sin\left(\frac{x}{2}\right) = 0\)
For \( \cos\left(\frac{3x}{2}\right) = 0 \), the solution is:
\(\frac{3x}{2} = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad x = \frac{\pi}{3} + \frac{2n\pi}{3}, \quad n \in \mathbb{Z}\)
For \( \sin\left(\frac{x}{2}\right) = 0 \), the solution is:
\(\frac{x}{2} = n\pi \quad \Rightarrow \quad x = 2n\pi, \quad n \in \mathbb{Z}\)
Thus, the general solution is the union of both sets:
\(x = 2n\pi \quad \text{or} \quad x = \frac{\pi}{3} + \frac{2n\pi}{3}, \quad n \in \mathbb{Z}\)
|
Q 1. The function \(f(x)=|\sin 4 x|+|\cos 2 x|\), is a periodic function with period; |
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(a) \(2 \pi\); |
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(b) \(\pi\); |
|
(c) \(\frac{\pi}{2}\); |
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(d) \(\frac{\pi}{4}\); |
|
Q 1. The sides of a triangle are \(\sin \alpha, \cos \alpha\) and \(\sqrt{1+\sin \alpha \cos \alpha}\) for some \(0<\alpha<\frac{\pi}{2}\) Then greatest angle of the triangle is; |
|
(a) \(60^{\circ}\); |
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(b) \(90^{\circ}\); |
|
(c) \(120^{\circ}\); |
|
(d) \(150^{\circ}\); |
|
Q 2. In a triangle \(A B C, 2 c a \sin \frac{A-B+C}{2}\) is equal to; |
|
(a) \(a^{2}+b^{2}-c^{2}\); |
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(b) \(c^{2}+a^{2}-b^{2}\); |
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(c) \(b^{2}-c^{2}-a^{2}\); |
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(d) \(c^{2}-a^{2}-b^{2}\); |
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Q 3. In a \(\triangle P Q R\), if \(3 \sin P+4 \cos Q=6\) and \(4 \sin Q+3 \cos P=1\), then the angle \(R\) is equal to; |
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(a) \(\frac{5 \pi}{6}\); |
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(b) \(\frac{\pi}{6}\); |
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(c) \(\frac{\pi}{4}\); |
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(d) \(\frac{3 \pi}{4}\); |
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Q 1. Number of solutions of the equation \(\tan x+\sec x=\) \(2 \cos x\) lying in the interval \([0, \pi]\) is; |
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(a) 0; |
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(b) 1; |
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(c) 2; |
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(d) 3; |
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Q 2. Find \(x\) from the equation: \(\operatorname{cosec}\left(90^{\circ}+\theta\right)+x \cos \theta \cot \left(90^{\circ}+\theta\right)=\sin \left(90^{\circ}+\theta\right)\); |
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(a) \(\cot \theta\); |
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(b) \(\tan \theta\); |
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(c) \(-\tan \theta\); |
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(d) \(-\cot \theta\); |
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Q 3. The most general value of \(\theta\) satisfying the equations \(\sin \theta=\sin \alpha\) and \(\cos \theta=\cos \alpha\) is; |
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(a) \(2 \mathrm{n} \pi+\alpha\); |
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(b) \(2 \mathrm{n} \pi-\alpha\); |
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(c) \(\mathrm{n} \pi+\alpha\); |
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(d) \(\mathrm{n} \pi-\alpha\); |
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Q 4. Number of solutions of equation, \(\sin 5 x \cos 3 x=\sin 6 x \cos 2 x\), in the interval \([0, \pi]\) is; |
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(a) 4; |
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(b) 5; |
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(c) 3; |
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(d) 2; |
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Q 5. The most general value of \(\theta\) satisfying the equation \(\cos \theta=\frac{1}{\sqrt{2}}\) and \(\tan \theta=-1\) is; |
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(a) \(2 \mathrm{n} \pi-7 \frac{\pi}{4}\); |
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(b) \(\mathrm{n} \pi-\frac{\pi}{4}\); |
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(c) \(\mathrm{n} \pi+\frac{\pi}{2}\); |
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(d) \(2 \mathrm{n} \pi+\frac{7 \pi}{4}\); |
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Q 6. The solution of the equation \(\cos ^{2} \theta+\sin \theta+1=0\), lies in the interval; |
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(a) \(\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\); |
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(b) \(\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)\); |
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(c) \(\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)\); |
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(d) \(\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)\); |
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Q 7. The number of values of \(x\) in the interval \([0,3 \pi]\) satisfying the equation \(2 \sin ^{2} x+5 \sin x-3=0\) is; |
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(a) 4; |
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(b) 6; |
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(c) 1; |
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(d) 2; |
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Q 8. The solution of the equation \([\sin x+\cos x]^{1+\sin 2 x}=2,-\pi \leq x \leq \pi\) is; |
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(a) \(\frac{\pi}{2}\); |
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(b) \(\pi\); |
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(c) \(\frac{\pi}{4}\); |
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(d) \(\frac{3 \pi}{4}\); |
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Q 9. If \(\mathrm{n}\) is any integer, then the general solution of the equation \(\cos x-\sin x=\frac{1}{\sqrt{2}}\) is; |
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(a) \(x=2 \mathrm{n} \pi-\frac{\pi}{12}\) or \(x=2 \mathrm{n} \pi+\frac{7 \pi}{12}\); |
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(b) \(x=\mathrm{n} \pi \pm \frac{\pi}{12}\); |
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(c) \(x=2 \mathrm{n} \pi+\frac{\pi}{12}\) or \(x=2 \mathrm{n} \pi-\frac{7 \pi}{12}\); |
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(d) \(x=\mathrm{n} \pi+\frac{\pi}{12}\) or \(x=\mathrm{n} \pi-\frac{7 \pi}{12}\); |
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Q 10. The number of values of \(x\) in the interval \([0,3 \pi]\) satisfying the equation \(2 \sin ^{2} x+5 \sin x-3=0\) is; |
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(a) 4; |
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(b) 6; |
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(c) 1; |
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(d) 2; |
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Q 1. A square of side \(a\) lies above the \(x\)-axis and has one vertex at the origin The side passing through the origin makes an angle \(\alpha\left(0<\alpha<\frac{\pi}{4}\right)\) with the positive direction of \(x\)-axis The equation of its diagonal not passing through the origin is; |
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(a) \(y(\cos \alpha-\sin \alpha)-x(\sin \alpha-\cos \alpha)=a\); |
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(b) \(y(\cos \alpha+\sin \alpha)+x(\sin \alpha-\cos \alpha)=a\); |
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(c) \(y(\cos \alpha+\sin \alpha)+x(\sin \alpha+\cos \alpha)=a\); |
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(d) \(y(\cos \alpha+\sin \alpha)+x(\cos \alpha-\sin \alpha)=a\); |
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Q 1. If an angle \(A\) of a \(\triangle A B C\) satisfies \(5 \cos A+3=0\), then the roots of the quadratic equation, \(9 x^{2}+27 x+20=0\) are; |
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(a) \(\sec A, \cot A\); |
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(b) \(\sin A, \sec A\); |
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(c) \(\sec A, \tan A\); |
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(d) \(\tan A, \cos A\); |
Trigonometric Inequalities
Trigonometric inequalities involve finding the range of values for an unknown angle that satisfies an inequality involving trigonometric functions. These inequalities are similar to algebraic inequalities but require special attention due to the periodic nature of trigonometric functions. Solving them often involves using trigonometric identities and understanding the behavior of trigonometric functions over specific intervals.
1. General Approach to Solving Trigonometric Inequalities
The basic steps to solve a trigonometric inequality are:
1. Isolate the trigonometric function: If possible, reduce the inequality to the form \( \sin(x) > k \), \( \cos(x) < k \), etc.
2. Find critical points: Solve the corresponding equality to determine the boundary or critical points where the trigonometric function takes the given value.
3. Use periodicity: Consider the periodic nature of the trigonometric function to determine all possible solutions in a given range (often \( (0, 2\pi) \)).
4. Test intervals: Based on the solutions obtained, test intervals within one period of the function to check if the inequality holds.
2. Common Trigonometric Inequalities
Here are some of the most frequently encountered trigonometric inequalities:
- Sine Inequality: \( \sin(x) > k \), \( \sin(x) < k \)
- Cosine Inequality: \( \cos(x) > k \), \( \cos(x) < k \)
- Tangent Inequality: \( \tan(x) > k \), \( \tan(x) < k \)
- Cosecant, Secant, and Cotangent Inequalities: These are less common but can be tackled similarly by transforming them into sine, cosine, or tangent inequalities using reciprocal identities.
3. Solving Trigonometric Inequalities
Let’s break down some common cases:
a. Inequalities Involving Sine and Cosine
Since both sine and cosine functions oscillate between -1 and 1, trigonometric inequalities involving sine and cosine often have restricted solutions based on the value of \( k \).
- Example 1: Solve \( \sin(x) > \frac{1}{2} \) for \( 0 \leq x \leq 2\pi \).
1. Step 1: Solve the Equality: First, solve \( \sin(x) = \frac{1}{2} \). The solutions in one period \( (0, 2\pi) \) are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).
2. Step 2: Test Intervals: The sine function is positive in the intervals \( \left(0, \pi\right) \), so for \( \sin(x) > \frac{1}{2} \), the solution lies between \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \).
3. Final Solution: The solution is \( x \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right) \).
- Example 2: Solve \( \cos(x) \leq -\frac{1}{2} \) for \( 0 \leq x \leq 2\pi \).
1. Step 1: Solve the Equality: First, solve \( \cos(x) = -\frac{1}{2} \). The solutions in one period are \( x = \frac{2\pi}{3} \) and \( x = \frac{4\pi}{3} \).
2. Step 2: Test Intervals: The cosine function is negative between \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \), and therefore, for \( \cos(x) \leq -\frac{1}{2} \), the solution is between \( \frac{2\pi}{3} \) and \( \frac{4\pi}{3} \).
3. Final Solution: The solution is \( x \in \left(\frac{2\pi}{3}, \frac{4\pi}{3}\right) \).
b. Inequalities Involving Tangent and Cotangent
The tangent and cotangent functions have a period of \( \pi \), meaning their values repeat every \( \pi \) units.
- Example 3: Solve \( \tan(x) < 1 \) for \( 0 \leq x \leq 2\pi \).
1. Step 1: Solve the Equality: First, solve \( \tan(x) = 1 \), which gives \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).
2. Step 2: Test Intervals: The tangent function is increasing in the intervals \( \left(0, \frac{\pi}{2}\right) \) and \( \left(\pi, \frac{3\pi}{2}\right) \). Since \( \tan(x) = 1 \) at \( \frac{\pi}{4} \) and \( \frac{5\pi}{4} \), for \( \tan(x) < 1 \), the solution lies in the intervals \( \left(0, \frac{\pi}{4}\right) \) and \( \left(\pi, \frac{5\pi}{4}\right) \).
3. Final Solution: The solution is \( x \in \left(0, \frac{\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \).
4. Special Cases
Some inequalities may involve the reciprocal trigonometric functions such as secant (\(\sec(x)\)), cosecant (\(\csc(x)\)), or cotangent (\(\cot(x)\)). In such cases, we can rewrite the inequality using the corresponding sine, cosine, or tangent functions.
- Example 4: Solve \( \sec(x) > 2 \) for \( 0 \leq x \leq 2\pi \).
1. Step 1: Rewrite in Terms of Cosine: The secant function is the reciprocal of cosine, so the inequality becomes \( \frac{1}{\cos(x)} > 2 \), or equivalently \( \cos(x) < \frac{1}{2} \).
2. Step 2: Solve the Equality: First, solve \( \cos(x) = \frac{1}{2} \), which gives solutions \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \).
3. Step 3: Test Intervals: Since cosine is less than \( \frac{1}{2} \) between \( \left(\frac{\pi}{2}, \frac{5\pi}{6}\right) \) and \( \left(\frac{7\pi}{6}, \frac{3\pi}{2}\right) \), the solution to the inequality is in those intervals.
4. Final Solution: The solution is \( x \in \left( \frac{\pi}{3}, \frac{5\pi}{3} \right) \).
5. Graphical Approach to Solving Trigonometric Inequalities
In some cases, graphing the trigonometric function can help visualize the solution. By plotting the function over one or more periods, you can easily see where the function lies above or below certain values and identify the intervals that satisfy the inequality.
6. Multiple Period Solutions
Since trigonometric functions are periodic, it is essential to extend solutions across multiple periods. For example, if you solve an inequality for \( 0 \leq x \leq 2\pi \), and the problem asks for all solutions, the general solution will include the periodicity term.
- Example: For \( \sin(x) > \frac{1}{2} \), the general solution can be written as:
\(
x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi, \quad n \in \mathbb{Z}
\)
Example Problem
Q1: Solve \( \sin(x) \geq -\frac{\sqrt{3}}{2} \) for \( 0 \leq x \leq 2\pi \).
- Solution:
1. First, solve the equality \( \sin(x) = -\frac{\sqrt{3}}{2} \). The solutions are \( x = \frac{4\pi}{3} \) and \( x = \frac{5\pi}{3} \).
2. Now, consider where \( \sin(x) \geq -\frac{\sqrt{3}}{2} \). The sine function is greater than or equal to \( -\frac{\sqrt{3}}{2} \) for \( x \in (0, \frac{4\pi}{3}) \cup (\frac{5\pi}{3}, 2\pi) \).
3. Final solution: \( x \in (0, \frac{4\pi}{3}) \cup (\frac{5
Key Concept: Trigonometric Identities, Hyperbolic Functions
Hyperbolic functions are analogs of trigonometric functions but are related to hyperbolas instead of circles. These functions are widely used in advanced mathematics, physics, and engineering, particularly in the study of differential equations, calculus, and complex analysis. Just like trigonometric identities, hyperbolic identities are crucial for simplifying expressions and solving equations involving hyperbolic functions.
1. Definitions of Hyperbolic Functions
The six main hyperbolic functions are:
- Hyperbolic Sine (\(\sinh(x)\)):
\(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
- Hyperbolic Cosine (\(\cosh(x)\)):
\(\cosh(x) = \frac{e^x + e^{-x}}{2}\)
- Hyperbolic Tangent (\(\tanh(x)\)):
\(\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}}\)
- Hyperbolic Cotangent (\(\coth(x)\)):
\(\coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{e^x + e^{-x}}{e^x - e^{-x}}\)
- Hyperbolic Secant (\(\sech(x)\)):
\(\sech(x) = \frac{1}{\cosh(x)}\)
- Hyperbolic Cosecant (\(\csch(x)\)):
\(\csch(x) = \frac{1}{\sinh(x)}\)
2. Basic Hyperbolic Identities
Just like with trigonometric functions, hyperbolic functions satisfy several important identities.
- Pythagorean-like Identity:
\(\cosh^2(x) - \sinh^2(x) = 1\)
This identity is analogous to the Pythagorean identity for trigonometric functions, but with a subtraction instead of addition.
- Other Identities:
\(1 - \tanh^2(x) = \sech^2(x)\)
\(\coth^2(x) - 1 = \csch^2(x)\)
3. Even and Odd Properties of Hyperbolic Functions
The hyperbolic functions exhibit even or odd properties, similar to trigonometric functions:
- Even Functions:
\(\cosh(-x) = \cosh(x)\)
- Odd Functions:
\(\sinh(-x) = -\sinh(x)\)
\(\tanh(-x) = -\tanh(x)\)
4. Hyperbolic Function Graphs
The graphs of hyperbolic functions are distinct from trigonometric functions, as hyperbolic functions grow exponentially for large positive and negative values of \(x\). Here are some important characteristics of the graphs:
- \(\sinh(x)\): The graph of the hyperbolic sine function is similar to a stretched-out sine curve but grows exponentially as \(x \to \infty\) and \(x \to -\infty\).
- \(\cosh(x)\): The graph of hyperbolic cosine is symmetric and resembles a "U" shape, with its minimum value occurring at \(x = 0\). It also grows exponentially for large positive and negative values of \(x\).
5. Inverse Hyperbolic Functions
Each of the hyperbolic functions has an inverse, which is useful for solving equations that involve hyperbolic functions. The inverse hyperbolic functions are:
- Inverse Hyperbolic Sine (\(\sinh^{-1}(x)\)):
\(\sinh^{-1}(x) = \ln\left(x + \sqrt{x^2 + 1}\right)\)
- Inverse Hyperbolic Cosine (\(\cosh^{-1}(x)\)):
\(\cosh^{-1}(x) = \ln\left(x + \sqrt{x^2 - 1}\right) \quad (x \geq 1)\)
- Inverse Hyperbolic Tangent (\(\tanh^{-1}(x)\)):
\(\tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right) \quad (-1 < x < 1)\)
6. Solving Hyperbolic Equations
Just like trigonometric equations, hyperbolic equations can be solved by applying the definitions and identities of hyperbolic functions.
- Example: Solve the equation \(\cosh(x) = 3\).
- Using the definition of \(\cosh(x)\), we have:
\(\cosh(x) = \frac{e^x + e^{-x}}{2} = 3\)
- Multiply by 2:
\(e^x + e^{-x} = 6\)
- Multiply the equation by \(e^x\):
\(e^{2x} + 1 = 6e^x\)
- Rearrange the equation into a quadratic form:
\(e^{2x} - 6e^x + 1 = 0\)
- Solve using the quadratic formula:
\(e^x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}\)
- Therefore, \(x = \ln(3 + 2\sqrt{2})\).
7. Application in IIT JEE
Hyperbolic functions are less frequently tested in IIT JEE than trigonometric functions, but they do appear in advanced topics, especially in calculus and complex analysis. Hyperbolic identities and equations are important in physics problems involving hyperbolic motion and in solving differential equations.
Example Problem
Q1: Solve the equation \( \tanh(x) = \frac{1}{2} \).
- Solution:
\(\tanh(x) = \frac{1}{2} = \frac{\sinh(x)}{\cosh(x)}\)
Using the identity \( 1 - \tanh^2(x) = \sech^2(x) \), we have:
\(1 - \left(\frac{1}{2}\right)^2 = \sech^2(x) = \frac{3}{4}\)
\(\cosh(x) = \frac{2}{\sqrt{3}}, \quad \sinh(x) = \frac{\cosh(x)}{2} = \frac{1}{\sqrt{3}}\)
Therefore:
\(x = \sinh^{-1}\left(\frac{1}{\sqrt{3}}\right) = \ln\left(\frac{1}{\sqrt{3}} + \sqrt{\frac{1}{3} + 1}\right)\)
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Q 1. Q5. The set of all the values of \(x\) satisfying the inequality \(\left(\cot ^{-1} x\right)^{2}-7\left(\cot ^{-1} x\right)+10>0\) is; |
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(A) \((-\infty, \cot 5) \cup(\cot 4, \cot 2)\); |
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(B) \((\cot 5, \cot 4)\); |
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(C) \((-\infty, \cot 5) \cup(\cot 2, \infty)\); |
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(D) \((\cot 2, \infty)\); |
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Q 2. Q5. The set of all the values of \(x\) satisfying the inequality \(\left(\cot ^{-1} x\right)^{2}-7\left(\cot ^{-1} x\right)+10>0\) is; |
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(A) \((-\infty, \cot 5) \cup(\cot 4, \cot 2)\); |
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(B) \((\cot 5, \cot 4)\); |
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(C) \((-\infty, \cot 5) \cup(\cot 2, \infty)\); |
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(D) \((\cot 2, \infty)\); |
Angles and Their Measures |
Properties of Trigonometric Functions |
Solution of Triangles |
Trigonometric Equations |
Trigonometric Equations, Angles and Their Measures |
Trigonometric Equations, Solution of Triangles, Angles and Their Measures |
Trigonometric Identities |
Trigonometric Identities, Angles and Their Measures |
Trigonometric Identities, Hyperbolic Functions |
Trigonometric Identities, Periodic Functions |
Trigonometric Identities, Solution of Triangles, Angles and Their Measures |
Trigonometric Identities, Trigonometric Equations |
Trigonometric Identities, Trigonometric Equations, Angles and Their Measures |
Trigonometric Identities, Trigonometric Equations, Solution of Triangles, Angles and Their Measures |
Trigonometric Inequalities |