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JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Trigonometric Ratios and Identities

If

sin θ + cos θ = \frac{1}{2}

, then

16 (sin (2 θ) + cos (4 θ) + sin (6 θ))

is equal to :

(A)27
(B)-23✓ Correct
(C)-27
(D)23

Answer: (B)

Solution

Squaring

sin θ + cos θ = \frac{1}{2}

gives

1 + sin 2 θ = \frac{1}{4}

, so

sin 2 θ = - \frac{3}{4}

.
Then

cos 4 θ = 1 - 2 sin^{2} 2 θ = 1 - \frac{9}{8} = - \frac{1}{8}

, and

sin 6 θ = 3 sin 2 θ - 4 sin^{3} 2 θ = 3 (- \frac{3}{4}) - 4 (- \frac{27}{64}) = - \frac{9}{4} + \frac{27}{16} = - \frac{9}{16}

Adding the three terms,

sin 2 θ + cos 4 θ + sin 6 θ = - \frac{3}{4} - \frac{1}{8} - \frac{9}{16} = - \frac{23}{16}

so

16 (sin 2 θ + cos 4 θ + sin 6 θ) = - 23

.

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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