How do you express sin(pi/ 8 ) * cos(( ( pi) / 3 ) sin(π8)cos((π3) without using products of trigonometric functions?

1 Answer
Apr 11, 2016

(sin((11pi)/24)-sin((5pi)/24))/2sin(11π24)sin(5π24)2

Explanation:

Use
sin 2A - sin 2B = 2 cos(A+B)sin(A-B)sin2Asin2B=2cos(A+B)sin(AB).

Here, A+B=pi/3 and A-B=pi/8A+B=π3andAB=π8.

So, 2A=11pi/24 and 2B=5pi/242A=11π24and2B=5π24.

Answer: (sin((11pi)/24)-sin((5pi)/24))/2sin(11π24)sin(5π24)2