Question

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

Answer 1

It may be "cheating", but I would just substitute `1/2` for `cos(\pi/3)`.

Explanation:

You're probably supposed to use the identity

`cos a sin b = (1/2)(sin(a+b)-sin(a-b))`.

Put in `a=\pi/3={8\pi}/24, b={5\pi}/8={15\pi}/24`.

Then

`cos(\pi/3)sin({5*pi}/8)=(1/2)(sin({23*\pi}/24)-sin({-7*pi}/24))`

`=(1/2)(sin({\pi}/24)+sin({7*\pi}/24))`

where in the last line we use `sin(\pi-x)=sin(x)` and `sin(-x)=-sin(x)`.

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

Answer 2

`- (1/2)cos (pi/8)`

Explanation:

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