# 8(cos(15π/8)sin(5π/8)) i would like to know the complete solution. thanks

Supposing that you need to conver the product into a sum, than you should use the formula:

`sin alpha*cos beta = (1/2)*[sin(alpha - beta) + sin(alpha+beta)]`

Hence, substituting `5pi/8`  for `alpha`  and `15pi/8`  for `beta`  yields:

`sin (5pi/8)*cos (15pi/8) = (1/2)*[sin(5pi/8 - 15pi/8) + sin(5pi/8 + 15pi/8)]`

Multiplying by 8 yields:

`8sin (5pi/8)*cos (15pi/8) = (8/2)*[sin(5pi/8 - 15pi/8) + sin(5pi/8 + 15pi/8)]`

`8sin (5pi/8)*cos (15pi/8) = 4*[sin(-10pi/8) + sin(20pi/8)]`

You need to remember that `sin(-alpha) = -sin alpha` , hence `sin(-10pi/8) = -sin(10pi/8) = -sin (5pi/4)`

Notice that `sin(5pi/4) = sin(pi/4 + pi) = sin(pi/4)*cos pi + sin pi*cos(pi/4) = -sqrt2/2 (cos pi=-1 and sin pi=0)` .

Hence, `-sin(5pi/4) = sqrt2/2.`

Evaluating `sin 20pi/8`  yields:

`sin 20pi/8 = sin 5pi/2 = sin(pi/2 + 4pi/2) = sin(pi/2 + 2pi) = sin pi/2 = 1` .

You need to substitute 1 for `sin 20pi/8`  and `sqrt2/2`  for `-sin(5pi/4)`  such that:

`8sin (5pi/8)*cos (15pi/8) = 4*(sqrt2/2 + 1)`

`8sin (5pi/8)*cos (15pi/8) = 2*(sqrt2 + 2)`

Hence, evaluating the product `8sin (5pi/8)*cos (15pi/8) ` yields `8sin (5pi/8)*cos (15pi/8) = 2*(sqrt2 + 2).`

Approved by eNotes Editorial Team