Question

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

• A. $27$
• B. $-23$
• C. $-27$
• D. $23$

Answer 1

The correct option is B $-23$

Given: $\sin \theta +\cos \theta =\frac{1}{2}$
squaring on both sides, we get
$\sin 2\theta =\frac{1}{4}-1$
$\therefore \sin 2\theta =-\frac{3}{4}$
and $\cos 4\theta =1-2(\sin 2\theta {)}^2$
$\Rightarrow \cos 4\theta =1-\frac{9}{8}$
$\therefore \cos 4\theta =-\frac{1}{8}$
and $\sin 6\theta =3\sin 2\theta -4{\sin }^{3}2\theta$
$\Rightarrow \sin 6\theta =-\frac{9}{4}+\frac{27}{16}$
$\therefore \sin 6\theta =-\frac{9}{16}$
Now,
$16\left(\sin \right(2\theta )+\cos (4\theta )+\sin (6\theta \left)\right)$
$=16(-\frac{3}{4}-\frac{1}{8}-\frac{9}{16})$
$=-23$