How do you express sin(pi/ 6 ) * cos( (3 pi) / 8 ) sin(π6)cos(3π8) without using products of trigonometric functions?

1 Answer
Mar 17, 2016

sqrt(2 - sqrt2)/4224

Explanation:

P = sin (pi/6).cos ((3pi)/8).P=sin(π6).cos(3π8).
Trig table --> sin pi/6 = 1/2sinπ6=12
Find cos ((3pi)/8)cos(3π8) by the identity: cos 2a = 2cos^2 a - 1cos2a=2cos2a1
cos ((6pi)/8) = 2cos^2 ((3pi)/8) - 1.cos(6π8)=2cos2(3π8)1.
cos ((3pi)/4) = -sqrt2/2 = 2cos^2 ((3pi)/8) - 1cos(3π4)=22=2cos2(3π8)1
2cos^2 ((3pi)/8) = 1 - sqrt2/2 = (2 - sqrt2)/22cos2(3π8)=122=222
cos^2 ((3pi)/8) = (2 - sqrt2)/4cos2(3π8)=224
cos ((3pi)/8) = sqrt(2 - sqrt2)/2cos(3π8)=222 (since cos ((3pi)/8)cos(3π8) is positive.
Finally,

P = (1/2)(sqrt(2 - sqrt2)/2) = (sqrt(2 - sqrt2)/4)P=(12)(222)=(224)