JEE Mathematics Trigonometric Functions MCQs Set 12

Answer: (d) 3

Explanation:
1. \( \tan\theta=\cot5\theta \Rightarrow \cos5\theta\cos\theta-\sin5\theta\sin\theta=0 \Rightarrow \cos6\theta=0 \).
2. \( \sin2\theta=1-2\sin^22\theta \Rightarrow (2\sin2\theta-1)(\sin2\theta+1)=0 \Rightarrow \sin2\theta=\frac12 \) or \( -1 \), giving \( \theta=\frac\pi{12},\frac{5\pi}{12},-\frac\pi4 \) in the interval.
3. For each of these \( 6\theta=\frac\pi2,\frac{5\pi}2,-\frac{3\pi}2 \), so cos 6 theta = 0 holds and none equals a multiple of pi/5. So there are 3 values.

Teacher's Note:
1. Reduce the first equation to cos 6 theta = 0.
2. Solve the second equation as a quadratic in sin 2 theta.

Answer: (c) no solution

Explanation:
1. \( \sin x-\sin3x=-2\cos2x\sin x \) and \( 2\sin2x=4\sin x\cos x \), so the left side is \( \sin x\,(2+4\cos x-4\cos^2x) \).
2. The bracket is \( 3-(2\cos x-1)^2\le3 \) and \( \sin x\le1 \), so the left side is at most 3.
3. Equality needs \( \sin x=1 \) (then \( \cos x=0 \) and the bracket is 2) together with the bracket equal to 3 (then \( \cos x=\frac12 \)): impossible. So the left side is always below 3 and there is no solution.

Teacher's Note:
1. Factor out sin x and complete the square in cos x.
2. Both factors would have to reach their maxima at the same x.

Answer: (b) 8

Explanation:
1. \( \cos^4x+\sin^4x=1-\frac12\sin^22x \) and \( \cos^6x+\sin^6x=1-\frac34\sin^22x \).
2. The equation becomes \( \frac54\cos^22x+2-\frac54\sin^22x=2 \Rightarrow \cos4x=0 \).
3. \( 4x=\frac\pi2+k\pi,\ 4x\in[0,8\pi] \) gives \( k=0,1,\ldots,7 \): 8 solutions.

Teacher's Note:
1. Use the standard sums of fourth and sixth powers.
2. Count the solutions of cos 4x = 0 in a range of length 8 pi for 4x.

Answer: (d) \( 2(\sqrt3-1) \)

Explanation:
1. Use \( \frac1{\sin a\sin b}=\frac{\cot a-\cot b}{\sin(b-a)} \) with \( b-a=\frac\pi6 \) and \( \sin\frac\pi6=\frac12 \).
2. The sum telescopes to \( 2\left[\cot\frac\pi4-\cot\left(\frac\pi4+\frac{13\pi}6\right)\right] \).
3. \( \frac\pi4+\frac{13\pi}6=2\pi+\frac{5\pi}{12} \) and \( \cot\frac{5\pi}{12}=2-\sqrt3 \), so the sum is \( 2\left[1-(2-\sqrt3)\right]=2(\sqrt3-1) \).

Teacher's Note:
1. Insert cot a - cot b = sin(b - a)/(sin a sin b) to telescope.
2. cot 75 degrees = 2 - sqrt3.

Answer: (a) 0.5

Explanation:
1. Write the left side as \( R\cos(x-\varphi) \) with \( \tan\varphi=\frac{2b}{\sqrt3a} \). The two roots are symmetric about \( \varphi \): \( \alpha+\beta=2\varphi \).
2. \( 2\varphi=\frac\pi3 \Rightarrow \varphi=\frac\pi6 \Rightarrow \frac{2b}{\sqrt3a}=\frac1{\sqrt3} \).
3. So \( \frac ba=\frac12 \).

Teacher's Note:
1. Both roots satisfy cos(x - phi) = c/R, so they lie symmetrically about phi.
2. Use tan phi = (coefficient of sin)/(coefficient of cos).

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. Let \( \theta=\frac\pi{n+2} \). The numerator is \( \frac12\sum_{k=0}^n\left[\cos\theta-\cos(2k+3)\theta\right] \). Since \( (n+2)\theta=\pi \), \( \sum_{k=0}^n\cos(2k+3)\theta=\frac{\sin(n+1)\theta\cos(n+3)\theta}{\sin\theta}=-\cos\theta \), so the numerator is \( \frac{n+2}2\cos\theta \).
2. Similarly the denominator \( \sum\sin^2(k+1)\theta=\frac{n+1}2-\frac12\sum_{k=1}^{n+1}\cos2k\theta=\frac{n+1}2+\frac12=\frac{n+2}2 \). Hence \( f(n)=\cos\frac\pi{n+2} \).
3. (i) \( f(4)=\cos\frac\pi6=\frac{\sqrt3}2 \): true. (ii) \( \cos^{-1}f(6)=\frac\pi8 \), \( \alpha=\tan\frac\pi8=\sqrt2-1 \) satisfies \( \alpha^2+2\alpha-1=0 \): true. (iii) \( \cos^{-1}f(5)=\frac\pi7 \), so \( \sin\pi=0 \): true. (iv) \( f(n)\to\cos0=1 \), so false.

Teacher's Note:
1. Reduce both sums with product-to-sum formulas and the identity (n + 2) theta = pi.
2. Evaluate each statement from the closed form f(n) = cos(pi/(n + 2)).

Answer: (c) 0

Explanation:
1. Multiply by \( \sin x\cos x \): \( \sqrt3\sin x+\cos x+2(\sin^2x-\cos^2x)=0 \Rightarrow 2\sin\left(x+\frac\pi6\right)=2\cos2x \).
2. \( \sin\left(x+\frac\pi6\right)=\cos2x=\sin\left(\frac\pi2-2x\right) \Rightarrow x+\frac\pi6=\frac\pi2-2x+2k\pi \) or \( x+\frac\pi6=\pi-\left(\frac\pi2-2x\right)+2k\pi \).
3. First: \( x=\frac\pi9+\frac{2k\pi}3 \): \( \frac\pi9,\frac{7\pi}9,-\frac{5\pi}9 \). Second: \( x=-\frac\pi3-2k\pi \): \( -\frac\pi3 \). Sum \( =\frac{\pi+7\pi-5\pi}9-\frac\pi3=0 \).

Teacher's Note:
1. Clear the denominators and write the left side in the form sin(x + pi/6) - cos 2x.
2. Collect the solutions from the two cases of sin A = sin B.

Answer: (b) \( \frac34\le A\le1 \)

Explanation:
1. With \( c=\cos^2x \): \( A=(1-c)+c^2=\left(c-\frac12\right)^2+\frac34 \).
2. For \( c\in[0,1] \): minimum \( \frac34 \) at \( c=\frac12 \), maximum 1 at \( c=0,1 \).

Teacher's Note:
1. Replace sin^2 x by 1 - cos^2 x.
2. Complete the square.

Answer: (b) \( \frac\pi6 \)

Explanation:
1. Square and add: \( 9+16+24(\sin P\cos Q+\cos P\sin Q)=37 \Rightarrow \sin(P+Q)=\frac12 \).
2. So \( P+Q=\frac\pi6 \) or \( \frac{5\pi}6 \). If \( P+Q=\frac\pi6 \) then both are below 30 degrees and \( 3\sin P+4\cos Q \lt 1.5+4 \lt 6 \), which is impossible.
3. Hence \( P+Q=\frac{5\pi}6 \) and \( R=\frac\pi6 \).

Teacher's Note:
1. Squaring and adding produces sin(P + Q).
2. Reject the case that makes the first equation unreachable.

Answer: (d) \( \sec A\operatorname{cosec}A+1 \)

Explanation:
1. \( \frac{\tan A}{1-\cot A}=\frac{\tan^2A}{\tan A-1} \) and \( \frac{\cot A}{1-\tan A}=\frac{-1}{\tan A(\tan A-1)} \).
2. Sum \( =\frac{\tan^3A-1}{\tan A(\tan A-1)}=\frac{\tan^2A+\tan A+1}{\tan A}=\tan A+1+\cot A \).
3. \( \tan A+\cot A=\sec A\operatorname{cosec}A \), so the expression is \( \sec A\operatorname{cosec}A+1 \).

Teacher's Note:
1. Write everything in terms of tan A and factor tan^3 A - 1.
2. tan A + cot A = 1/(sin A cos A).

Answer: (d) \( \sqrt3:1 \)

Explanation:
1. With tower height \( h \): the distances from the foot are \( h\sqrt3,\ h,\ \frac h{\sqrt3} \) for A, B, C.
2. \( AB=h(\sqrt3-1) \) and \( BC=h\left(1-\frac1{\sqrt3}\right)=\frac{h(\sqrt3-1)}{\sqrt3} \), so \( AB:BC=\sqrt3:1 \).

Teacher's Note:
1. Express each horizontal distance as h cot(angle).
2. The factor (sqrt3 - 1) cancels.

Answer: (c) 5

Explanation:
1. With pillar height \( h \): distance at A is \( h\sqrt3 \) and at B is \( \frac h{\sqrt3} \), so \( AB=\frac{2h}{\sqrt3} \) is covered in 10 minutes.
2. Remaining distance \( \frac h{\sqrt3} \) is half of AB, so it takes 5 minutes.

Teacher's Note:
1. Speed is constant, so time is proportional to distance.
2. Compare the remaining distance with AB.

Answer: (a) 7

Explanation:
1. \( \cos x+\cos3x=2\cos2x\cos x \) and \( \cos2x+\cos4x=2\cos3x\cos x \), so the left side is \( 2\cos x(\cos2x+\cos3x)=4\cos x\cos\frac{5x}2\cos\frac x2 \).
2. \( \cos x=0 \): \( \frac\pi2,\frac{3\pi}2 \). \( \cos\frac{5x}2=0 \): \( \frac\pi5,\frac{3\pi}5,\pi,\frac{7\pi}5,\frac{9\pi}5 \). \( \cos\frac x2=0 \): \( x=\pi \) (already counted).
3. Distinct solutions: 2 + 5 = 7.

Teacher's Note:
1. Group cos x + cos 3x and cos 2x + cos 4x.
2. Do not count x = pi twice.

Answer: (b) \( \frac29 \)

Explanation:
1. Take \( AB=1,\ AP=2 \): \( \tan\angle APB=\frac12 \) and \( \tan\angle APC=\frac14 \).
2. \( \tan\beta=\frac{\frac12-\frac14}{1+\frac18}=\frac29 \).

Teacher's Note:
1. beta is the difference between two elevation angles at P.
2. Use the tangent subtraction formula.

Answer: (c) \( \frac{13}9 \)

Explanation:
1. \( \cos\left(\frac\pi6+x\right)\cos\left(\frac\pi6-x\right)=\cos^2x-\sin^2\frac\pi6=\cos^2x-\frac14 \), so the bracket is \( \cos^2x-\frac34 \).
2. \( 8\cos x\left(\cos^2x-\frac34\right)=8\cos^3x-6\cos x=2\cos3x=1 \Rightarrow \cos3x=\frac12 \).
3. \( 3x=\frac\pi3,\frac{5\pi}3,\frac{7\pi}3 \) for \( x\in[0,\pi] \), so \( x=\frac\pi9,\frac{5\pi}9,\frac{7\pi}9 \) and the sum is \( \frac{13\pi}9 \).

Teacher's Note:
1. Use cos(a + x)cos(a - x) = cos^2 x - sin^2 a.
2. Recognise 4cos^3 x - 3cos x.

Answer: (b) 100

Explanation:
1. Let \( M \) be the foot of the tower (the mid-point of QR) and h the height. Then \( PM=h\cot45^\circ=h \) and \( QM=h\cot30^\circ=h\sqrt3 \).
2. Since \( PQ=PR \), \( PM\perp QR \): \( PM^2+QM^2=PQ^2 \Rightarrow h^2+3h^2=200^2 \Rightarrow h=100 \) m.

Teacher's Note:
1. The median of an isosceles triangle from the apex is perpendicular to the base.
2. Apply Pythagoras in triangle PMQ.

Answer: (a) \( \frac1{512} \)

Explanation:
1. With \( x=\frac\pi{2^{10}} \): the cosines are \( \cos2^8x,\cos2^7x,\ldots,\cos x \) (nine factors).
2. \( \cos x\cos2x\cdots\cos2^8x=\frac{\sin2^9x}{2^9\sin x} \) and \( 2^9x=\frac\pi2 \).
3. Multiplying by \( \sin x \): the value is \( \frac{\sin\frac\pi2}{2^9}=\frac1{512} \).

Teacher's Note:
1. Apply the doubling product formula for cosines.
2. The final factor sin(pi/2) equals 1.

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

Explanation:
1. By AM-GM, \( \sin^4\alpha+4\cos^4\beta+1+1\ge4\sqrt[4]{4\sin^4\alpha\cos^4\beta}=4\sqrt2|\sin\alpha\cos\beta| \). Equality holds, so \( \sin^4\alpha=1 \) and \( 4\cos^4\beta=1 \) with \( \sin\alpha\cos\beta \gt 0 \).
2. \( \sin\alpha=1,\ \cos\beta=\frac1{\sqrt2} \), so \( \sin\beta=\frac1{\sqrt2} \) for \( \beta\in[0,\pi] \).
3. \( \cos(\alpha+\beta)-\cos(\alpha-\beta)=-2\sin\alpha\sin\beta=-\sqrt2 \).

Teacher's Note:
1. Treat the left side as a sum of four terms and apply AM-GM.
2. Equality in AM-GM fixes sin alpha and cos beta.

Answer: (a) 5

Explanation:
1. The left side is at least 1 and the right side is at most 1, so equality needs \( \sin x=0 \) and \( \cos^23x=1 \).
2. \( \sin x=0 \Rightarrow x=k\pi \), and then \( \cos3k\pi=\pm1 \) automatically.
3. \( k\pi\in\left[-\frac{5\pi}2,\frac{5\pi}2\right] \) gives \( k=-2,-1,0,1,2 \): 5 solutions.

Teacher's Note:
1. Compare the ranges of the two sides.
2. Both conditions are satisfied simultaneously at multiples of pi.

Answer: (a) \( \sin x=|\sin y| \)

Explanation:
1. \( \sqrt{\sin^2x-2\sin x+5}=\sqrt{(\sin x-1)^2+4}\ge2 \), so the first factor is at least 4.
2. \( 4^{-\sin^2y}\ge\frac14 \). The product is at least 1, so the inequality holds only when equality occurs: \( \sin x=1 \) and \( \sin^2y=1 \).
3. Then \( \sin x=1=|\sin y| \).

Teacher's Note:
1. Bound each factor separately.
2. The product can only equal 1 when both bounds are tight.

Answer: (d) 16

Explanation:
1. For poles of heights \( a \) and \( b \), the height of the intersection is \( \frac{ab}{a+b} \) (similar triangles).
2. \( \frac{20\cdot80}{100}=16 \) m.

Teacher's Note:
1. The result does not depend on the distance between the poles.
2. The harmonic-type formula ab/(a + b) applies.

Answer: (a) 10

Explanation:
1. Rewrite: \( (k+1)t^2-\sqrt2\lambda t-(1-k)=0 \): \( \tan\alpha+\tan\beta=\frac{\sqrt2\lambda}{k+1} \) and \( \tan\alpha\tan\beta=-\frac{1-k}{k+1} \).
2. \( \tan(\alpha+\beta)=\frac{\sqrt2\lambda/(k+1)}{1+\frac{1-k}{k+1}}=\frac{\sqrt2\lambda}2=\frac\lambda{\sqrt2} \).
3. \( \tan^2(\alpha+\beta)=\frac{\lambda^2}2=50 \Rightarrow \lambda=\pm10 \).

Teacher's Note:
1. Use the sum and product of the roots.
2. The factor k + 1 cancels in tan(alpha + beta).

Answer: (a) 8

Explanation:
1. \( \log_{1/2}|\sin x|+\log_{1/2}|\cos x|=2 \Rightarrow \log_{1/2}|\sin x\cos x|=2 \Rightarrow |\sin x\cos x|=\frac14 \).
2. \( |\sin2x|=\frac12 \). In \( [0,2\pi] \), \( 2x\in[0,4\pi] \) and \( \sin2x=\pm\frac12 \) each have 4 solutions.
3. All 8 have sin x and cos x non-zero, so they are valid.

Teacher's Note:
1. Combine the logarithms of the same base.
2. sin 2x = 1/2 and sin 2x = -1/2 each have four solutions over two full periods.