JEE Mathematics Trigonometric Functions MCQs Set 17

Answer: (c) \(rac35\)

Explanation:
1. \(\cos\theta=\frac7{25}\).
2. \(\sin\frac\theta2=\sqrt{\dfrac{1-\frac7{25}}2}=\sqrt{\dfrac9{25}}=\dfrac35\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) \(2\cos3\theta\)

Explanation:
1. \(x=e^{\pm i\theta}\), so \(x^3+x^{-3}=2\cos3\theta\).
2. Check: \((2\cos\theta)^3-3(2\cos\theta)=2(4\cos^3\theta-3\cos\theta)=2\cos3\theta\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(2n\pi+\frac\pi2\)

Explanation:
1. \(\sin\theta+\cos\theta=1\) gives \(\theta=2n\pi\) or \(2n\pi+\frac\pi2\).
2. \(\theta=2n\pi\) makes \(\cot\theta\) undefined, so \(\theta=2n\pi+\frac\pi2\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (b) \(n\pi\pm\frac\pi3\)

Explanation:
1. \(\sin(x+\alpha)\sin(x-\alpha)=\sin^2x-\sin^2\alpha\) and \(\sin3\alpha=\sin\alpha(3-4\sin^2\alpha)\).
2. So \(3-4\sin^2\alpha=4\sin^2x-4\sin^2\alpha\), giving \(\sin^2x=\frac34\), \(x=n\pi\pm\frac\pi3\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) \(n\pi+\frac\pi4\)

Explanation:
1. \(\tan\theta+\cot\theta=\dfrac2{\sin2\theta}=2\) gives \(\sin2\theta=1\).
2. \(2\theta=2n\pi+\frac\pi2\), so \(\theta=n\pi+\frac\pi4\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (b) \(\theta=(3n+1)\frac\pi9,\ n\in I\)

Explanation:
1. \(\tan\theta+\tan2\theta=\sqrt3(1-\tan\theta\tan2\theta)\), so \(\tan3\theta=\sqrt3\).
2. \(3\theta=n\pi+\frac\pi3\), \(\theta=(3n+1)\frac\pi9\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(2n\pi+\alpha\)

Explanation:
1. Both sine and cosine agree only when the angles differ by a multiple of \(2\pi\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(n\pi\pm\frac\pi6\)

Explanation:
1. \(\dfrac{2\tan^2\theta}{1-\tan^2\theta}=1\) gives \(\tan^2\theta=\frac13\).
2. \(\theta=n\pi\pm\frac\pi6\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) \(30^\circ,90^\circ,150^\circ\)

Explanation:
1. \(2\sin^2x-3\sin x+1=0\) gives \(\sin x=1\) or \(\frac12\).
2. \(x=90^\circ\), \(30^\circ\), \(150^\circ\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(2n\pi\pm\frac\pi3\)

Explanation:
1. \(\dfrac1{\sin\theta\cos\theta}=\dfrac2{\sin\theta}\) gives \(\cos\theta=\frac12\).
2. \(\theta=2n\pi\pm\frac\pi3\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) \(\left(n+\frac16\right)\frac\pi5\)

Explanation:
1. \(\dfrac{\tan2\theta+\tan3\theta}{1-\tan2\theta\tan3\theta}=\dfrac1{\sqrt3}\), so \(\tan5\theta=\dfrac1{\sqrt3}\).
2. \(5\theta=n\pi+\frac\pi6\), so \(\theta=\left(n+\frac16\right)\frac\pi5\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (b) 1 radian

Explanation:
1. By definition, an arc equal to the radius subtends an angle of one radian.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(\frac{121\pi}{540}\)

Explanation:
1. \(40^\circ20'=\frac{121}3{}^\circ\), and \(\frac{121}3\times\frac\pi{180}=\frac{121\pi}{540}\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) \(180^\circ\)

Explanation:
1. \(\pi\) radian \(=180^\circ\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) \(\frac{\sqrt3}2\)

Explanation:
1. \(\frac{31\pi}3=10\pi+\frac\pi3\), so \(\sin\frac{31\pi}3=\sin\frac\pi3=\frac{\sqrt3}2\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (b) \(\frac16\)

Explanation:
1. \(\sin\theta=-\frac45\): \(\mathrm{cosec}\,\theta+\cot\theta=-\frac54+\frac34=-\frac12\).
2. \(\sec\theta-\tan\theta=-\frac53-\frac43=-3\), so the ratio is \(\frac16\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) \(-\sqrt2\sin x\)

Explanation:
1. \(\cos P-\cos Q=-2\sin\frac{P+Q}2\sin\frac{P-Q}2=-2\sin\frac{3\pi}4\sin x=-\sqrt2\sin x\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) \(2\sin x\)

Explanation:
1. \(\sin x-\sin3x=-2\cos2x\sin x\) and \(\sin^2x-\cos^2x=-\cos2x\).
2. The ratio is \(2\sin x\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(2\sec\theta\)

Explanation:
1. \(\frac1x=\sec\theta-\tan\theta\), so the sum is \(2\sec\theta\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) \(2\)

Explanation:
1. \(\tan70^\circ-\tan20^\circ=\tan50^\circ(1+\tan70^\circ\tan20^\circ)=2\tan50^\circ\) since \(\tan70^\circ\tan20^\circ=1\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) \(4\)

Explanation:
1. \(=\dfrac{\cos10^\circ-\sqrt3\sin10^\circ}{\sin10^\circ\cos10^\circ}=\dfrac{2\cos70^\circ}{\frac12\sin20^\circ}=\dfrac{2\sin20^\circ}{\frac12\sin20^\circ}=4\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (c) \(4\cos x\cos2x\cos3x\)

Explanation:
1. \(1+\cos2x=2\cos^2x\) and \(\cos4x+\cos6x=2\cos5x\cos x\).
2. Total \(=2\cos x(\cos x+\cos5x)=4\cos x\cos2x\cos3x\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (a) \(n\pi+(-1)^n\frac\pi4+\frac\pi3\)

Explanation:
1. \(\sin\theta-\sqrt3\cos\theta=\sqrt2\), i.e. \(2\sin\left(\theta-\frac\pi3\right)=\sqrt2\).
2. \(\theta-\frac\pi3=n\pi+(-1)^n\frac\pi4\).

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (b) \(2\)

Explanation:
1. \(\cot54^\circ=\tan36^\circ\) and \(\cot70^\circ=\tan20^\circ\), so each ratio is 1.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.

Answer: (d) I and II both are true

Explanation:
1. \((\Sigma\cos)^2+(\Sigma\sin)^2=3+2\Sigma\cos(\beta-\gamma)=0\) exactly when the sum equals \(-\frac32\).
2. A sum of two squares is zero only if each is zero, so both I and II hold.

Teacher's Note:
1. Reduce every angle to an acute angle in the first quadrant using the ASTC rule, then evaluate.
2. For general solutions use sin x = sin a gives \( x = n \pi + (-1)^n a, \cos x = \cos a \) gives x = 2 n pi +- a, tan x = tan a gives x = n pi + a.