How do you express #sin(pi/ 6 ) * cos( ( 15 pi) / 8 ) # without using products of trigonometric functions?

1 Answer
Jan 18, 2016

#-(1/4)sqrt(2 + sqrt2)#

Explanation:

Trig Table of Special arcs --> #sin (pi/6) = 1/2#
#cos ((15pi)/8) = cos (-pi/8 + 2pi) = -cos (pi/8)#
Find (cos (pi/8)) by using trig identity:
#cos (pi/4) = sqrt2/2 = 2cos^2 (pi/8) - 1#
#2cos^2 (pi/8) = 1 + sqrt2/2 = (2 + sqrt2)/2#
#cos^2 (pi/8) = (2 + sqrt2)/4.#
#cos (pi/8) = + sqrt(2 + sqrt2)/2#. (because cos (pi/8) is positive)
Finally,
#sin (pi/6).cos ((15pi)/8) = -1/2cos (pi/8) =#
#= -(1/4)sqrt(2 + sqrt2)#