# How do you express cos(pi/ 3 ) * sin( ( 5 pi) / 8 )  without using products of trigonometric functions?

Aug 1, 2016

It may be "cheating", but I would just substitute $\frac{1}{2}$ for $\cos \left(\setminus \frac{\pi}{3}\right)$.

#### Explanation:

You're probably supposed to use the identity

$\cos a \sin b = \left(\frac{1}{2}\right) \left(\sin \left(a + b\right) - \sin \left(a - b\right)\right)$.

Put in $a = \setminus \frac{\pi}{3} = \frac{8 \setminus \pi}{24} , b = \frac{5 \setminus \pi}{8} = \frac{15 \setminus \pi}{24}$.

Then

$\cos \left(\setminus \frac{\pi}{3}\right) \sin \left(\frac{5 \cdot \pi}{8}\right) = \left(\frac{1}{2}\right) \left(\sin \left(\frac{23 \cdot \setminus \pi}{24}\right) - \sin \left(\frac{- 7 \cdot \pi}{24}\right)\right)$

$= \left(\frac{1}{2}\right) \left(\sin \left(\frac{\setminus \pi}{24}\right) + \sin \left(\frac{7 \cdot \setminus \pi}{24}\right)\right)$

where in the last line we use $\sin \left(\setminus \pi - x\right) = \sin \left(x\right)$ and $\sin \left(- x\right) = - \sin \left(x\right)$.

As you can see, this is unwieldy compared with just putting in $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$. The trigonometric product-sum and product-difference relations are more useful when you can't evaluate either factor in the product.

Aug 2, 2016

$- \left(\frac{1}{2}\right) \cos \left(\frac{\pi}{8}\right)$

#### Explanation:

$P = \cos \left(\frac{\pi}{3}\right) . \sin \left(\frac{5 \pi}{8}\right)$
Trig table --> $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$
Trig unit circle and property of complementary arcs -->
$\sin \left(\frac{5 \pi}{8}\right) = \sin \left(\frac{\pi}{8} + \frac{4 \pi}{8}\right) = \sin \left(\frac{\pi}{8} + \frac{\pi}{2}\right) =$
$= - \cos \left(\frac{\pi}{8}\right) .$
P can be expressed as:
$P = - \left(\frac{1}{2}\right) \cos \left(\frac{\pi}{8}\right)$
NOTE. We can evaluate $\cos \left(\frac{\pi}{8}\right)$ by using the trig identity:
$1 + \cos \left(\frac{\pi}{4}\right) = 2 {\cos}^{2} \left(\frac{\pi}{8}\right)$