Question

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

• A. 23
• B. -27
• C. -23
• D. 27

Answer 1

Answer: (C)

$\sin \theta +\cos \theta =\frac{1}{2}$
${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =\frac{1}{4}$
$\sin 2\theta =-\frac{3}{4}$
Now: $\cos 4\theta =1-2{\sin }^{2}2\theta$
$=1-2{\left(-\frac{3}{4}\right)}^2$
$=1-2\times \frac{9}{16}=-\frac{1}{8}$
$\sin 6\theta =3\sin 2\theta -4{\sin }^{3}2\theta$
$=\left(3-4{\sin }^{2}2\theta \right)\cdot \sin 2\theta$
$=\left[3-4\left(\frac{9}{16}\right)\right]\cdot \left(-\frac{3}{4}\right)$
$\Rightarrow \left[\frac{3}{4}\right]\times \left(-\frac{3}{4}\right)=-\frac{9}{16}$
$16[\sin 2\theta +\cos 4\theta +\sin 6\theta ]$
$16\left(-\frac{3}{4}-\frac{1}{8}-\frac{9}{16}\right)=-23$