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3.Trigonometrical Ratios, Functions and Identities
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If $\sin \theta+\cos \theta=\frac{1}{2}$, then $16(\sin (2 \theta)+\cos (4 \theta)+\sin (6 \theta))$ is equal to:
A
$27$
B
$-27$
C
$-23$
D
$23$
(JEE MAIN-2021)
Solution
$16[2 \sin 4 \theta \cos 2 \theta+\cos 4 \theta]$
$16\left[4 \sin 2 \theta \cos ^{2} 2 \theta+2 \cos ^{2} 2 \theta-1\right]$
Now:
$\sin \theta+\cos \theta=\frac{1}{2}$
$1+\sin 2 \theta=\frac{1}{4}$
$\sin 2 \theta=-\frac{3}{4}$
$\cos ^{2} 2 \theta=1-\frac{9}{16}=\frac{7}{16}$
$16\left[-4(-3 / 4) \times \frac{7}{16}+2 \times \frac{7}{16}-1\right]$
$16\left[\frac{-7}{16}-1\right] \Rightarrow-23$
Standard 11
Mathematics
