How do you express sin(pi/12) * cos(pi/8 ) sin(π12)cos(π8) without products of trigonometric functions?

1 Answer
Dec 21, 2016

(1/2)[sin (pi/24) + (1/2)sin ((5pi)/24)](12)[sin(π24)+(12)sin(5π24)]

Explanation:

P = sin (pi/12).cos (pi/8)P=sin(π12).cos(π8)
Use trig identity:
sin a.cos b = (1/2)[sin (a - b) + (1/2)(sin (a + b)]sina.cosb=(12)[sin(ab)+(12)(sin(a+b)]
We have:
sin (a - b) = sin ((pi/8) - (pi/12)) = sin (pi/24)sin(ab)=sin((π8)(π12))=sin(π24)
sin (a + b) = sin ((pi/8) + (pi/12)) = sin ((5pi)/24)sin(a+b)=sin((π8)+(π12))=sin(5π24)
There for, P can be expressed as:
P = (1/2)[sin (pi/24) + (1/2)sin ((5pi)/24)]P=(12)[sin(π24)+(12)sin(5π24)]