# How do you express sin(pi/12) * cos(( pi)/2 )  without products of trigonometric functions?

$\textcolor{b l u e}{\sin \left(\frac{\pi}{12}\right) \cdot \cos \left(\frac{\pi}{2}\right) = \frac{1}{2} \sin \left(\frac{7 \pi}{12}\right) - \frac{1}{2} \sin \left(\frac{5 \pi}{12}\right)}$

#### Explanation:

Let us use the sum and difference formulas

$\sin \left(x + y\right) = \sin x \cdot \cos y + \cos x \cdot \sin y$

$\sin \left(x - y\right) = \sin x \cdot \cos y - \cos x \cdot \sin y$

After addition of equal values, we have

$\sin \left(x + y\right) + \sin \left(x - y\right) = 2 \cdot \sin x \cdot \cos y$

and we have the formula

$\sin x \cdot \cos y = \frac{1}{2} \sin \left(x + y\right) + \frac{1}{2} \sin \left(x - y\right)$

We can now use the given

$\sin \left(\frac{\pi}{12}\right) \cdot \cos \left(\frac{\pi}{2}\right) = \frac{1}{2} \sin \left(\frac{\pi}{12} + \frac{\pi}{2}\right) + \frac{1}{2} \sin \left(\frac{\pi}{12} - \frac{\pi}{2}\right)$

$\sin \left(\frac{\pi}{12}\right) \cdot \cos \left(\frac{\pi}{2}\right) = \frac{1}{2} \sin \left(\frac{7 \pi}{12}\right) + \frac{1}{2} \sin \left(\frac{- 5 \pi}{12}\right)$

$\sin \left(\frac{\pi}{12}\right) \cdot \cos \left(\frac{\pi}{2}\right) = \frac{1}{2} \sin \left(\frac{7 \pi}{12}\right) - \frac{1}{2} \sin \left(\frac{5 \pi}{12}\right)$

God bless....I hope the explanation is useful.