JEE Mathematics Trigonometric Functions MCQs Set 06

Answer: (c) 4

Explanation:
1. \( \sin x+\sin^3x=1-\sin^2x=\cos^2x \), i.e. \( \sin x(1+\sin^2x)=\cos^2x \).
2. Square: \( \sin^2x(1+\sin^2x)^2=\cos^4x \). With \( c=\cos^2x \): \( (1-c)(2-c)^2=c^2 \).
3. Expand: \( (1-c)(4-4c+c^2)=4-8c+5c^2-c^3=c^2 \), so \( c^3-4c^2+8c-4=0 \).
4. Hence \( \cos^6x-4\cos^4x+8\cos^2x=c^3-4c^2+8c=4 \).

Teacher's Note:
1. Isolate cos^2 x from the given relation and square to remove the odd powers of sine.
2. Let c = cos^2 x and recognise the target expression in the resulting cubic.

Answer: (a) 0

Explanation:
1. \( \cos(\sin x)=\sin\left(\tfrac\pi2-\sin x\right) \), so the equation requires \( \cos x=\tfrac\pi2-\sin x+2k\pi \) or \( \cos x=\tfrac\pi2+\sin x+2k\pi \).
2. That is \( \sin x+\cos x=\tfrac\pi2+2k\pi \) or \( \cos x-\sin x=\tfrac\pi2+2k\pi \). Both left sides have absolute value at most \( \sqrt2\approx1.414 \), while \( \tfrac\pi2\approx1.571 \) and other values of the right side are even further out.
3. So there is no solution.

Teacher's Note:
1. Rewrite the cosine as a sine of the complementary angle and compare the arguments.
2. Compare with the maximum value root 2 of sin x + cos x.

Answer: (c) 13

Explanation:
1. Let \( c=\cos x \), so \( 3c^2=c+1 \) and \( \cos2x=2c^2-1=\tfrac{2c-1}3 \).
2. The two values of c are \( \tfrac{1\pm\sqrt{13}}6 \), so \( t=\cos2x=\tfrac{-2\pm\sqrt{13}}9 \).
3. Sum \( t_1+t_2=-\tfrac49 \), product \( t_1t_2=\tfrac{4-13}{81}=-\tfrac19 \), so \( 9t^2+4t-1=0 \).
4. Comparing with \( at^2+bt-1=0 \): \( a=9,\ b=4 \), \( a+b=13 \).

Teacher's Note:
1. Express cos 2x in terms of cos x and use the given relation to reduce it to a linear form.
2. Form the quadratic with the two values of cos 2x as roots.

Answer: (d) a straight line passing through the point \( \left(\tfrac\pi2,-\sin^21\right) \) and parallel to the x-axis

Explanation:
1. \( \cos x\cos(x+2)=\tfrac12[\cos(2x+2)+\cos2] \) and \( \cos^2(x+1)=\tfrac12[1+\cos(2x+2)] \).
2. Subtract: \( y=\tfrac12(\cos2-1)=-\sin^21 \), a constant.
3. So the graph is the horizontal line \( y=-\sin^21 \), which passes through \( (\tfrac\pi2,-\sin^21) \).

Teacher's Note:
1. Use the product-to-sum formula for the first term and the half-angle form for the second.
2. A constant function has a horizontal line as graph.

Answer: (a) 28

Explanation:
1. Write \( \cot(5\pi\sin\theta)=\tan\left(\tfrac\pi2-5\pi\sin\theta\right) \). Then \( 5\pi\cos\theta=\tfrac\pi2-5\pi\sin\theta+k\pi \), i.e. \( \sin\theta+\cos\theta=\dfrac{2k+1}{10} \).
2. Since \( |\sin\theta+\cos\theta|\le\sqrt2\approx1.414 \), \( |2k+1|\le14 \): odd values \( -13,-11,\dots,13 \), i.e. 14 values.
3. Each value of magnitude below \( \sqrt2 \) is attained twice in \( (0,2\pi) \), so there are \( 14\times2=28 \) solutions.

Teacher's Note:
1. Convert the cotangent into a tangent of the complementary angle.
2. sin x + cos x = sqrt2 sin(x + pi/4) takes each value strictly inside (-sqrt2, sqrt2) twice per period.

Answer: (b) n = 1, 2, 4, 5

Explanation:
1. Let \( u=\sin\tfrac x2-\cos\tfrac x2 \). Then \( u^2=1-\sin x \), so the equation becomes \( u=u^2 \): \( u=0 \) or \( u=1 \).
2. The inequality gives \( -\tfrac{\pi}4\le\tfrac x2\le\tfrac{5\pi}4 \), i.e. \( -\tfrac\pi2\le x\le\tfrac{5\pi}2 \).
3. Case \( u=0 \): \( x=\tfrac\pi2+2k\pi \), in range: \( \tfrac\pi2,\tfrac{5\pi}2 \) (n = 1, 5).
4. Case \( u=1 \): \( \sin(\tfrac x2-\tfrac\pi4)=\tfrac1{\sqrt2} \) gives \( x=\pi+4k\pi \) or \( x=2\pi+4k\pi \), in range: \( \pi,\ 2\pi \) (n = 2, 4).
5. So n = 1, 2, 4, 5.

Teacher's Note:
1. Notice that (sin(x/2) - cos(x/2))^2 equals 1 - sin x.
2. Convert the inequality to a range for x and list the solutions in it.

Answer: (d) \( \tfrac{n\pi}3 \)

Explanation:
1. The relation \( \tan A+\tan B+\tan C=\tan A\tan B\tan C \) holds when \( A+B+C=m\pi \) (when all tangents exist). Here \( 6x=m\pi \), so \( x=\tfrac{m\pi}6 \).
2. If \( m \) is odd then \( 3x=\tfrac{m\pi}2 \) and \( \tan3x \) is undefined, so those values are rejected.
3. Remaining \( m=2n \): \( x=\tfrac{n\pi}3 \).

Teacher's Note:
1. The identity tan A + tan B + tan C = tan A tan B tan C follows from tan(A + B + C) = 0.
2. Reject values that make any of the tangents undefined.

Answer: (b) 1

Explanation:
1. \( y=2\cos\tfrac{\alpha+\beta}2\left[\cos\tfrac{\alpha-\beta}2-\cos\tfrac{\alpha+\beta}2\right]=(\cos\alpha+\cos\beta)-2\cos^2\tfrac{\alpha+\beta}2 \).
2. \( 2\cos^2\tfrac{\alpha+\beta}2=1+\cos(\alpha+\beta) \), so \( y=\cos\alpha+\cos\beta-1-\cos(\alpha+\beta) \).
3. Hence \( x-y=1 \).

Teacher's Note:
1. Use 2 sin P sin Q = cos(P - Q) - cos(P + Q) on the first two factors of y.
2. Use the double angle formula to remove the square.

Answer: (d) \( \tfrac{61}{24} \)

Explanation:
1. Write \( \csc^2\theta=1+\cot^2\theta \) and \( \sec^2\theta=1+\tan^2\theta \): \( f=\tfrac12+\tfrac23+\tfrac38+\tfrac23\cot^2\theta+\tfrac38\tan^2\theta \).
2. A.M.-G.M.: \( \tfrac23\cot^2\theta+\tfrac38\tan^2\theta\ge2\sqrt{\tfrac23\cdot\tfrac38}=1 \), equality at \( \tan^4\theta=\tfrac{16}9 \).
3. Minimum \( \tfrac12+\tfrac23+\tfrac38+1=\tfrac{61}{24} \).

Teacher's Note:
1. Convert csc^2 and sec^2 into cot^2 and tan^2 so that A.M.-G.M. gives a constant.
2. tan^2 x cot^2 x = 1, so the product of the two terms is constant.

Answer: (c) \( \tfrac12(\tan27x-\tan x) \)

Explanation:
1. \( \tan3x-\tan x=\dfrac{\sin2x}{\cos x\cos3x}=\dfrac{2\sin x}{\cos3x} \), so \( \dfrac{\sin x}{\cos3x}=\tfrac12(\tan3x-\tan x) \).
2. Similarly the other terms are \( \tfrac12(\tan9x-\tan3x) \) and \( \tfrac12(\tan27x-\tan9x) \).
3. The sum telescopes to \( \tfrac12(\tan27x-\tan x) \).

Teacher's Note:
1. Write tan 3x - tan x as sin 2x over cos x cos 3x.
2. Each term then becomes half of a difference of tangents.

Answer: (b) \( \text{independent of }\gamma\text{ and equal to }(\alpha+\beta)^2 \)

Explanation:
1. Expand: \( \alpha^2+\beta^2+\alpha\beta(\tan^2\gamma+\cot^2\gamma) \).
2. \( \tan^2\gamma+\cot^2\gamma=(\tan\gamma+\cot\gamma)^2-2=\dfrac4{\sin^22\gamma}-2 \) and \( 4\cot^22\gamma=\dfrac{4}{\sin^22\gamma}-4 \).
3. So the expression is \( \alpha^2+\beta^2+\alpha\beta(-2+4)=(\alpha+\beta)^2 \), free of gamma but dependent on alpha and beta.

Teacher's Note:
1. Expand the product and use tan + cot = 2/sin 2 gamma.
2. cot^2 2 gamma = 1/sin^2 2 gamma - 1.

Answer: (c) C and D

Explanation:
1. (C) At \( \theta=\tfrac{3\pi}2 \): \( \sin\tfrac{3\pi}2=-1 \) so the first term is 1; every later argument is a multiple of \( \pi \), so the other terms vanish. \( f_4=1 \) (true).
2. (D) At \( \theta=\pi \) every argument \( 2^k\pi \) has sine 0, so \( f_5(\pi)=0 \) (true).
3. (A) \( f_2(\tfrac\pi4)=\sin^4\tfrac\pi4+\tfrac14\sin^4\tfrac\pi2+\tfrac1{16}\sin^4\pi=\tfrac14+\tfrac14=\tfrac12\ne\tfrac1{\sqrt2} \) (false).
4. (B) \( f_3(\tfrac\pi8)=\sin^4\tfrac\pi8+\tfrac14\sin^4\tfrac\pi4+\tfrac1{16}\sin^4\tfrac\pi2+\tfrac1{64}\sin^4\pi\approx0.1464 \), not \( \tfrac{2+\sqrt2}4\approx0.854 \) (false).

Teacher's Note:
1. Substitute the special angles directly: the sines of multiples of pi are zero.
2. Evaluate each sum term by term.

Answer: (c) A, B and C

Explanation:
1. Put \( x=\cos\theta \): \( 4x^3-3x=\cos3\theta \), so the equation is \( \cos3\theta=-\tfrac{\sqrt3}2 \).
2. Then \( 3\theta=\tfrac{5\pi}6 \) or \( \tfrac{7\pi}6 \) (plus multiples of \( 2\pi \)), giving \( \theta=\tfrac{5\pi}{18},\ \tfrac{7\pi}{18},\ \tfrac{17\pi}{18} \) as the three distinct roots.
3. Check (D): \( \cos\tfrac{23\pi}{18}=-\cos\tfrac{5\pi}{18} \) would need \( 3\theta=\tfrac{23\pi}6 \) with cosine \( +\tfrac{\sqrt3}2 \), so it is not a root.

Teacher's Note:
1. Recognise 4x^3 - 3x as cos 3 theta with x = cos theta.
2. A cubic has at most three roots; find three distinct angles.

Answer: (d) all of \( a,\ b \) and \( \phi \)

Explanation:
1. Let \( u=\phi+b \) and \( v=\phi+a+b \), so \( v-u=a \).
2. Identity: \( \cos^2u+\cos^2v-2\cos(v-u)\cos u\cos v=\sin^2(v-u) \).
3. So the expression is \( \sin^2a+\cos^2a=1 \), independent of a, b and phi.

Teacher's Note:
1. Introduce u and v so that a = v - u.
2. The standard identity cos^2 u + cos^2 v - 2 cos(u - v) cos u cos v = sin^2(u - v) finishes the problem.

Answer: (a) 4950

Explanation:
1. \( \cos4x=2\cos^22x-1 \), so \( 2c^2-7c+5=0 \) with \( c=\cos2x \): \( c=1 \) or \( c=\tfrac52 \) (impossible).
2. So \( \cos2x=1 \), \( x=m\pi \).
3. In \( [0,314] \): \( m\pi\le314 \) gives \( m\le99 \) (\( 99\pi\approx311.0 \), \( 100\pi\approx314.16>314 \)).
4. Sum \( =\pi(0+1+\cdots+99)=4950\pi \).

Teacher's Note:
1. Convert cos 4x to a quadratic in cos 2x and discard the root larger than 1.
2. Check how many multiples of pi fit below 314.

Answer: (c) \( \tfrac54 \)

Explanation:
1. \( \tan18^\circ\tan72^\circ=1 \) and \( \tan108^\circ\tan162^\circ=(-\tan72^\circ)(-\tan18^\circ)=1 \), so the four tangents multiply to 1.
2. \( \sin36^\circ\sin144^\circ=\sin^236^\circ \), \( \cos54^\circ\cos126^\circ=-\cos^254^\circ=-\sin^236^\circ \) and \( \cos180^\circ=-1 \), so the product is \( \sin^436^\circ \).
3. With \( \sin36^\circ=2\sin18^\circ\cos18^\circ \): \( k=\dfrac{\sin^436^\circ}{\sin^218^\circ}=16\sin^218^\circ\cos^418^\circ \). With \( \sin18^\circ=\tfrac{\sqrt5-1}4 \) this equals \( \tfrac54 \).

Teacher's Note:
1. Pair complementary and supplementary angles to simplify the product.
2. sin 18 degrees equals (sqrt5 - 1)/4.

Answer: (a) \( \tfrac12 \)

Explanation:
1. Use \( \sin^3x=\tfrac{3\sin x-\sin3x}4 \) and \( \cos^3x=\tfrac{3\cos x+\cos3x}4 \).
2. The left side becomes \( \tfrac34(\sin x\cos3x+\cos x\sin3x)+\tfrac14(-\sin3x\cos3x+\cos3x\sin3x)=\tfrac34\sin4x \).
3. So \( \tfrac34\sin4x=\tfrac38 \) and \( \sin4x=\tfrac12 \).

Teacher's Note:
1. Replace sin^3 and cos^3 by their triple-angle expansions.
2. The extra terms cancel and sin x cos 3x + cos x sin 3x = sin 4x.

Answer: (d) \( \tfrac58 \)

Explanation:
1. \( \tan^2x+1=\sec^2x \), so \( f(x)=5\sin^3x\cos^3x=5\left(\tfrac{\sin2x}2\right)^3=\tfrac58\sin^32x \).
2. The maximum of \( \sin^32x \) is 1, so the maximum is \( \tfrac58 \).

Teacher's Note:
1. Replace tan^2 + 1 by sec^2 to simplify the denominator.
2. sin x cos x = (sin 2x)/2.

Answer: (d) 4

Explanation:
1. Group as squares: \( (3\cos x-1)^2+(\sec y-2)^2=0 \) since \( 1+4=5 \).
2. Both squares vanish: \( \cos x=\tfrac13 \) and \( \sec y=2\ (\cos y=\tfrac12) \).
3. In \( (0,2\pi) \) each has 2 solutions, so there are \( 2\times2=4 \) pairs.

Teacher's Note:
1. Complete the squares in cos x and sec y.
2. A sum of two squares is zero only if each square is zero.

Answer: (b) 2 and 3

Explanation:
1. The expression is largest when \( \sin x=-1 \): \( a^2+6-5a\le0 \).
2. So \( (a-2)(a-3)\le0 \) and \( 2\le a\le3 \).
3. Integral values: 2 and 3.

Teacher's Note:
1. Find the worst case over x: -6 sin x is largest when sin x = -1.
2. Solve the resulting quadratic inequality in a.