# How do you express sin(pi/ 4 ) * sin( ( pi) / 3 )  without using products of trigonometric functions?

May 20, 2016

$\sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{3}\right) = \frac{1}{2} \left[\cos \left(\frac{\pi}{12}\right) + \cos \left(\frac{5 \pi}{12}\right)\right]$

#### Explanation:

We know that $\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$ ..............(1)

and $\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$ ..............(2)

Now subtracting (2) from (1)

$2 \sin A \sin B = \cos \left(A - B\right) - \cos \left(A + B\right)$ or

$\sin A \sin B = \frac{1}{2} \cos \left(A - B\right) - \frac{1}{2} \cos \left(A + B\right)$

Hence, $\sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{3}\right) = \frac{1}{2} \cos \left(\left(\frac{\pi}{4}\right) - \left(\frac{\pi}{3}\right)\right) - \frac{1}{2} \cos \left(\left(\frac{\pi}{4}\right) + \left(\frac{\pi}{3}\right)\right)$

= $\frac{1}{2} \cos \left(- \frac{\pi}{12}\right) - \frac{1}{2} \cos \left(\frac{7 \pi}{12}\right)$,

but $\cos \left(- x\right) = \cos x$ and $\cos \left(\pi - x\right) = - \cos x$,

hence above is equal to

$\frac{1}{2} \cos \left(\frac{\pi}{12}\right) - \frac{1}{2} \cos \left(\pi - \frac{5 \pi}{12}\right)$

= $\frac{1}{2} \cos \left(\frac{\pi}{12}\right) + \frac{1}{2} \cos \left(\frac{5 \pi}{12}\right) = \frac{1}{2} \left[\cos \left(\frac{\pi}{12}\right) + \cos \left(\frac{5 \pi}{12}\right)\right]$