If sinθ+cosθ=1/2, then 16sin2θ+cos4θ+sin6θ is equal to:

Correct answer: (C) -23. sinθ+cosθ=1/2 sin^2θ+cos^2θ+2sinθcosθ=1/4⇒sin2θ=-3/4 Now: cos4θ=1-2sin^22θ =1-2-34^2 =1-2×9/16=-1/8 And sin6θ=3sin2θ-4sin^32θ =3-4sin^22θ·sin2θ =3-49/16·-3/4 =3/4×-3/4=-9/16 So, 16sin2θ+cos4θ+sin6θ =16-3/4-1/8-9/16=-23

Q39 / 90 · JEE Main 2021 (27 Jul Shift 1) · Mathematics

If $\sin θ + \cos θ = \frac{1}{2},$ then $16 (\sin (2 θ) + \cos (4 θ) + \sin (6 θ))$ is equal to:

A
$23$
B
$- 27$
Correct answer
C
$- 23$
D
$27$

Solution

$\sin θ + \cos θ = \frac{1}{2}$

$\sin^{2} θ + \cos^{2} θ + 2 \sin θ \cos θ = \frac{1}{4} \Rightarrow \sin 2 θ = - \frac{3}{4}$

Now:

$\cos 4 θ = 1 - 2 \sin^{2} 2 θ$

$= 1 - 2 {(- \frac{3}{4})}^{2}$

$= 1 - 2 \times \frac{9}{16} = - \frac{1}{8}$

And $\sin 6 θ = 3 \sin 2 θ - 4 \sin^{3} 2 θ$

$= (3 - 4 \sin^{2} 2 θ) \cdot \sin 2 θ$

$= [3 - 4 (\frac{9}{16})] \cdot (- \frac{3}{4})$

$= [\frac{3}{4}] \times (- \frac{3}{4}) = - \frac{9}{16}$

So, $16 [\sin 2 θ + \cos 4 θ + \sin 6 θ]$

$= 16 (- \frac{3}{4} - \frac{1}{8} - \frac{9}{16}) = - 23$

Hence, the correct answer is option: 'C'.

If sinθ+cosθ=12, then 16sin2θ+cos4θ+sin6θ is equal to:

Solution

Relevant Questions

If $\sin θ + \cos θ = \frac{1}{2},$ then $16 (\sin (2 θ) + \cos (4 θ) + \sin (6 θ))$ is equal to: