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

1 Answer
Feb 17, 2017

#sqrt(2 + sqrt2)/4#

Explanation:

#P = sin (pi/6).cos (pi/8)#
Trig table gives --> sin (pi/6) = 1/2
#P = (1/2)cos (pi/8)#.
We can evaluate #cos (pi/8)# by using trig identity:
#2cos^2 a = 1 + cos 2a#
In this case:
#2cos^2 (pi/8) = 1 + cos (pi/4) = 1 + sqrt2/2 = (2 + sqrt2)/2#
#cos^2 (pi/8) = (2 + sqrt2)/4#
#cos (pi/8) = +- sqrt(2 + sqrt2)/2#.
Since #cos (pi/8)# is positive, take the positive answer.
Finally,
#P = (1/2)cos (pi/8) = sqrt(2 + sqrt2)/4#