How do you express #sin(pi/ 8 ) * cos(( 7 pi)/6 ) # without using products of trigonometric functions?

1 Answer
Jan 24, 2016

#-(sqrt3)(sqrt(2 - sqrt2)/4)#

Explanation:

First, find #sin (pi/8)# and #cos ((7pi)/6)# separately.
Call #sin (pi/8) = sin t#
Use the trig identity: #cos (2t) = 1 - 2sin^2 t#
#cos (pi/4) = sqrt2/2 = 1 - sin^2 t#
#2sin^2 t = 1 - sqrt2 = (2 - sqrt2)/2#
#sin^2 t = (2 - sqrt2)/4#
#sin (pi/8) = sin t = (sqrt(2 - sqrt2))/2#

#cos ((7pi)/6) = cos (pi/6 + pi) = - cos (pi/6) = - sqrt3/2#

Finally, #sin (pi/8).cos ((7pi)/6) = - (sqrt3)(sqrt(2 - sqrt2)/4)#