JEE Mathematics Trigonometrical Functions and Identities MCQs Set 08

Answer: (a) \( 2\cos\dfrac\theta{2^n} \)

Explanation:
1. \( \sqrt{2(1+\cos\theta)}=\sqrt{4\cos^2\tfrac\theta2}=2\cos\tfrac\theta2 \) (one 2).
2. Then \( \sqrt{2+2\cos\tfrac\theta2}=2\cos\tfrac\theta4 \) (two 2's), and so on; the cosines are positive because the angles are below 90 degrees.
3. With n twos the result is \( 2\cos\dfrac\theta{2^n} \).

Teacher's Note:
1. Use 1 + cos x = 2 cos squared (x / 2) repeatedly from the innermost root outwards.

Answer: (b) \( \cot\dfrac\pi{16} \)

Explanation:
1. Use \( \cot x=\tan x+2\cot2x \).
2. \( \cot\tfrac\pi{16}=\tan\tfrac\pi{16}+2\cot\tfrac\pi8=\tan\tfrac\pi{16}+2\left(\tan\tfrac\pi8+2\cot\tfrac\pi4\right) \).
3. With \( \cot\tfrac\pi4=1 \), this is \( \tan\tfrac\pi{16}+2\tan\tfrac\pi8+4 \).

Teacher's Note:
1. The identity cot x - tan x = 2 cot 2x lets you build a chain of angles that doubles.

Answer: (b) 4

Explanation:
1. \( \dfrac{\sqrt3\cos20^\circ-\sin20^\circ}{\sin20^\circ\cos20^\circ}=\dfrac{2\sin(60^\circ-20^\circ)}{\tfrac12\sin40^\circ} \).
2. \( =\dfrac{2\sin40^\circ}{\tfrac12\sin40^\circ} \).
3. So the value is 4.

Teacher's Note:
1. Write root 3 cos x - sin x as 2 sin(60 degrees - x).

Answer: (c) 4

Explanation:
1. \( \cos\left(\tfrac\pi4-\theta\right)=\sin\left(\tfrac\pi2-\tfrac\pi4+\theta\right)=\sin\left(\tfrac\pi4+\theta\right) \).
2. The expression is \( 1+3\sin\left(\tfrac\pi4+\theta\right) \).
3. Its maximum is \( 1+3=4 \).

Teacher's Note:
1. Convert the cosine to a sine of the same angle using the complement.

Answer: (c) 4

Explanation:
1. \( \dfrac{\cos10^\circ-\sqrt3\sin10^\circ}{\sin10^\circ\cos10^\circ}=\dfrac{2\cos(10^\circ+60^\circ)}{\tfrac12\sin20^\circ} \).
2. \( \cos70^\circ=\sin20^\circ \), so the expression is \( \dfrac{2\sin20^\circ}{\tfrac12\sin20^\circ} \).
3. The value is 4.

Teacher's Note:
1. Write cos x - root 3 sin x as 2 cos(x + 60 degrees).

Answer: (c) \( \sqrt2 \)

Explanation:
1. HP gives \( \cos x=\dfrac{2\cos(x-y)\cos(x+y)}{\cos(x-y)+\cos(x+y)}=\dfrac{2(\cos^2x-\sin^2y)}{2\cos x\cos y} \), so \( \cos^2x\cos y=\cos^2x-\sin^2y \).
2. Hence \( \cos^2x(1-\cos y)=\sin^2y=(1-\cos y)(1+\cos y) \), and for \( y\ne0 \): \( \cos^2x=1+\cos y=2\cos^2\tfrac y2 \).
3. So \( \left|\cos x\sec\tfrac y2\right|=\sqrt2 \).

Teacher's Note:
1. Use the HP condition as the harmonic mean of the outer terms and the product-to-sum identities.

Answer: (c) HP

Explanation:
1. Expand \( \sin(\alpha+\gamma)=\sin\alpha\cos\gamma+\cos\alpha\sin\gamma \) and divide throughout by \( \sin\alpha\sin\beta\sin\gamma \).
2. This gives \( 2\cot\beta=\cot\gamma+\cot\alpha \), so \( \cot\alpha,\cot\beta,\cot\gamma \) are in AP.
3. Therefore \( \tan\alpha,\tan\beta,\tan\gamma \) are in HP.

Teacher's Note:
1. Divide by the product of sines to get cotangents.

Answer: (c) \( \tfrac12 \)

Explanation:
1. \( f(\theta)=\dfrac{\cos\theta}{\sin\theta+\cos\theta} \).
2. Put \( \beta=\tfrac{5\pi}4-\alpha \): \( \sin\beta+\cos\beta=\sqrt2\sin\left(\beta+\tfrac\pi4\right)=\sqrt2\sin\left(\tfrac{3\pi}2-\alpha\right)=-\sqrt2\cos\alpha \), and \( \cos\beta=\cos\left(\tfrac{5\pi}4-\alpha\right)=-\tfrac{1}{\sqrt2}(\cos\alpha+\sin\alpha) \).
3. Then \( f(\alpha)f(\beta)=\dfrac{\cos\alpha}{\sin\alpha+\cos\alpha}\cdot\dfrac{-\frac1{\sqrt2}(\cos\alpha+\sin\alpha)}{-\sqrt2\cos\alpha}=\dfrac12 \).

Teacher's Note:
1. Express f in terms of sine and cosine and substitute beta in terms of alpha.

Answer: (a) \( -\dfrac{627}{725} \)

Explanation:
1. Sum of roots \( =\tfrac{13}4 \) and product \( =\tfrac{15}8 \).
2. \( \tan\dfrac{\alpha+\beta}2=\dfrac{13/4}{1-15/8}=-\dfrac{26}7 \).
3. \( \cos(\alpha+\beta)=\dfrac{1-t^2}{1+t^2}=\dfrac{49-676}{49+676}=-\dfrac{627}{725} \).

Teacher's Note:
1. Use the tangent addition formula for the half angles, then the cosine in terms of tan of the half angle.

Answer: (a) 1

Explanation:
1. Since \( 0<\alpha+\beta<\tfrac\pi2 \) and \( 0\le\alpha-\beta<\tfrac\pi4 \), \( \sin(\alpha+\beta)=\tfrac45 \) and \( \sin(\alpha-\beta)=\tfrac35 \).
2. \( \sin2\alpha=\sin\left[(\alpha+\beta)+(\alpha-\beta)\right]=\tfrac45\cdot\tfrac45+\tfrac35\cdot\tfrac35 \).
3. \( =\tfrac{16+9}{25}=1 \).

Teacher's Note:
1. Write 2 alpha as the sum of the two given angles and fix the signs of the sines from the quadrant.

Answer: (b) \( \lambda \)

Explanation:
1. Multiply numerator and denominator of the given fraction by \( 1+\sin\alpha-\cos\alpha \): the denominator becomes \( (1+\sin\alpha)^2-\cos^2\alpha=2\sin\alpha(1+\sin\alpha) \).
2. So \( \lambda=\dfrac{2\sin\alpha(1+\sin\alpha-\cos\alpha)}{2\sin\alpha(1+\sin\alpha)} \).
3. This equals \( \dfrac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha} \).

Teacher's Note:
1. Multiply by the conjugate-type factor to use the difference of squares.

Answer: (c) \( \cot A \)

Explanation:
1. \( 1\pm\sin2A=(\cos A\pm\sin A)^2 \). Because \( |\tan A|<1 \) and A is acute (or its negative), \( \cos A+\sin A>0 \) and \( \cos A-\sin A>0 \).
2. So the roots are \( \cos A+\sin A \) and \( \cos A-\sin A \).
3. The ratio is \( \dfrac{2\cos A}{2\sin A}=\cot A \).

Teacher's Note:
1. Write one plus or minus sin 2A as a perfect square.

Answer: (b) \( \tan3\theta \)

Explanation:
1. \( \tan\left(\tfrac\pi3+\theta\right)\tan\left(\tfrac\pi3-\theta\right)=\dfrac{3-\tan^2\theta}{1-3\tan^2\theta} \).
2. Multiplying by \( \tan\theta \) gives \( \dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta} \).
3. This is \( \tan3\theta \).

Teacher's Note:
1. Use the tangent addition formulas and the triple-angle formula for tan.

Answer: (b) 3

Explanation:
1. With \( t=\tan\theta \): \( \tan\left(\theta+\tfrac\pi3\right)+\tan\left(\theta-\tfrac\pi3\right)=\dfrac{(t+\sqrt3)(1+\sqrt3t)+(t-\sqrt3)(1-\sqrt3t)}{1-3t^2}=\dfrac{8t}{1-3t^2} \).
2. Adding t: \( \dfrac{t(1-3t^2)+8t}{1-3t^2}=\dfrac{9t-3t^3}{1-3t^2}=3\cdot\dfrac{3t-t^3}{1-3t^2}=3\tan3\theta \).
3. So \( k=3 \).

Teacher's Note:
1. Combine the two shifted tangents over a common denominator and compare with the tan 3 theta formula.

Answer: (b) \( p\sqrt{2-p^2} \)

Explanation:
1. \( p^2=1-\sin40^\circ \), so \( \sin40^\circ=1-p^2 \) and \( 1+\sin40^\circ=2-p^2 \).
2. \( \cos40^\circ=(\cos20^\circ-\sin20^\circ)(\cos20^\circ+\sin20^\circ) \) and \( \cos20^\circ+\sin20^\circ=\sqrt{1+\sin40^\circ}>0 \).
3. Hence \( \cos40^\circ=p\sqrt{2-p^2} \).

Teacher's Note:
1. Factor cos 40 as the product of cos 20 - sin 20 and cos 20 + sin 20.

Answer: (c) \( xyz=x+y+z \)

Explanation:
1. The geometric sums give \( x=\dfrac1{\sin^2\phi},\ y=\dfrac1{\cos^2\phi},\ z=\dfrac1{1-\sin^2\phi\cos^2\phi} \).
2. \( x+y=\dfrac1{\sin^2\phi\cos^2\phi}=xy \), and \( z=\dfrac{xy}{xy-1} \), i.e. \( z(xy-1)=xy \).
3. So \( xyz=xy+z=x+y+z \).

Teacher's Note:
1. Sum each geometric series and note that x + y equals xy.

Answer: (a) \( \dfrac2{\sqrt5+1}\sqrt{10-2\sqrt5} \)

Explanation:
1. \( \tan x-\cot x=-2\cot2x \), so the value is \( -2\cot126^\circ=2\cot54^\circ=2\tan36^\circ \).
2. \( \tan36^\circ=\dfrac{\sin36^\circ}{\cos36^\circ}=\dfrac{\sqrt{10-2\sqrt5}/4}{(\sqrt5+1)/4}=\dfrac{\sqrt{10-2\sqrt5}}{\sqrt5+1} \).
3. So the value is \( \dfrac2{\sqrt5+1}\sqrt{10-2\sqrt5}\approx1.4531 \).

Teacher's Note:
1. Convert the difference to a single cotangent of a double angle and use the exact values of sin 36 and cos 36.

Answer: (c) \( \tfrac12\sqrt{5-\sqrt5} \)

Explanation:
1. \( \cos9^\circ-\sin9^\circ=\sqrt2\cos(9^\circ+45^\circ)=\sqrt2\cos54^\circ=\sqrt2\sin36^\circ \).
2. \( \sin36^\circ=\dfrac{\sqrt{10-2\sqrt5}}4 \), so the value is \( \dfrac{\sqrt{20-4\sqrt5}}4=\dfrac{2\sqrt{5-\sqrt5}}4 \).
3. That is \( \tfrac12\sqrt{5-\sqrt5} \).

Teacher's Note:
1. Write cos minus sin as root 2 times a single cosine.

Answer: (a) 0

Explanation:
1. With \( \theta=\tfrac\pi{10}=18^\circ \): \( 2\tan\theta+3\sec\theta-4\cos\theta=\dfrac{2\sin\theta+3-4\cos^2\theta}{\cos\theta} \).
2. \( 4\cos^2\theta=4-4\sin^2\theta \), so the numerator is \( 4\sin^2\theta+2\sin\theta-1 \).
3. \( \sin18^\circ=\tfrac{\sqrt5-1}4 \) satisfies \( 4s^2+2s-1=0 \), so the numerator is 0 and the value is 0.

Teacher's Note:
1. Reduce to a quadratic in sin 18 degrees and use its exact value.

Answer: (b) 0

Explanation:
1. \( \tan70^\circ=\cot20^\circ \), so \( \tan20^\circ-\tan70^\circ=\tan20^\circ-\cot20^\circ=-2\cot40^\circ \).
2. \( 2\tan50^\circ=2\cot40^\circ \).
3. The sum is \( -2\cot40^\circ+2\cot40^\circ=0 \).

Teacher's Note:
1. Use complementary angles and tan x - cot x = - 2 cot 2x.

Answer: (a) 1

Explanation:
1. \( k\alpha+(2n-k)\alpha=2n\alpha=\tfrac\pi2 \), so \( \cot k\alpha\cdot\cot(2n-k)\alpha=1 \) for each k.
2. The factors \( k=1,\dots,2n-1 \) pair up as \( (k,2n-k) \), leaving the middle factor \( \cot n\alpha=\cot\tfrac\pi4=1 \).
3. So the product is 1.

Teacher's Note:
1. Pair complementary angles whose sum is a right angle.

Answer: (a) \( \tfrac1{64} \)

Explanation:
1. \( \cos12^\circ\cos24^\circ\cos48^\circ\cos96^\circ=\dfrac{\sin192^\circ}{16\sin12^\circ}=-\tfrac1{16} \) and \( \cos84^\circ=-\cos96^\circ \), so \( \cos12^\circ\cos24^\circ\cos48^\circ\cos84^\circ=\tfrac1{16} \).
2. \( \cos36^\circ\cos72^\circ=\tfrac14 \).
3. The product is \( \tfrac1{16}\cdot\tfrac14=\tfrac1{64} \).

Teacher's Note:
1. Use the doubling product cos x cos 2x cos 4x cos 8x and the value cos 36 cos 72 = 1/4.

Answer: (c) \( \tfrac12 \)

Explanation:
1. Multiply by \( 2\sin\tfrac\pi{11} \): \( 2\sin\tfrac\pi{11}\cos\tfrac{(2k-1)\pi}{11}=\sin\tfrac{2k\pi}{11}-\sin\tfrac{(2k-2)\pi}{11} \).
2. The sum telescopes to \( \sin\tfrac{10\pi}{11}-\sin0=\sin\tfrac{\pi}{11} \).
3. So the sum is \( \dfrac{\sin(\pi/11)}{2\sin(\pi/11)}=\tfrac12 \).

Teacher's Note:
1. Multiply by 2 sin(pi/11) so that the products of sine and cosine telescope.

Answer: (b) 3

Explanation:
1. Since \( \sin^4x\le1 \), we need \( \cos^7x\ge0 \), so \( \cos x\ge0 \) and \( \cos^7x\le\cos^2x \).
2. Also \( \sin^4x\le\sin^2x \), so equality requires \( \cos x\in\{0,1\} \).
3. The solutions in \( [-\pi,\pi] \) are \( x=-\tfrac\pi2,\,0,\,\tfrac\pi2 \): 3 solutions.

Teacher's Note:
1. Show the left side is at most 1 and find the cases of equality.