Question

The expression $\sin {27}^o\cos {57}^o\sin {87}^o$ simplifies to

• A. $\frac{\sin 9^o}{4}$
• B. $\frac{\cos 9^o}{4}$
• C. $\frac{\sin 9^o}{2}$
• D. $\frac{\cos 9^o}{2}$

Answer 1

Answer: (B)

$\frac{\cos 9^o}{4}$

$\sin 27\cos 57\sin 87$ simplifies to

$\Rightarrow \cos 57\sin (90-3)$

$\Rightarrow \sin 27\cos 57\cos 3$

$\Rightarrow \sin 27\left[\frac{\cos (57+3)+\cos (57-3)}{2}\right]$

$\Rightarrow \frac{1}{2}\sin 27[\cos 60+\cos 54]=\frac{1}{2}\sin 27(\frac{1}{2}+\cos 54)$

Let $\sin 27=\theta$

$\Rightarrow \frac{1}{2}\sin \theta \left(\frac{1}{2}\cos 2\theta \right)$

$\Rightarrow \frac{1}{2}\sin \theta (\frac{1}{2}+1-{\sin }^2\theta )$

$\Rightarrow \frac{1}{2}\sin \theta (\frac{3}{2}-2{\sin }^2\theta )$

$\Rightarrow \frac{3}{4}\sin \theta -{\sin }^3\theta$

$\Rightarrow \frac{3\sin \theta -4{\sin }^3\theta }{4}$

$\Rightarrow \frac{\sin 3\theta }{4}$

$\Rightarrow \frac{\sin 3\left(27\right)}{4}$

$=\frac{\sin {81}^o}{4}$

$=\frac{\cos 9^o}{4}$.