How do you express #sin(pi/ 4 ) * sin( ( 5 pi) / 12 ) # without using products of trigonometric functions?

1 Answer
Jun 24, 2016

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

Explanation:

Product #P = sin (pi/4).sin ((5pi)/12)#
Trig table -->
#sin (pi/4) = sqrt2/2#
P can be expressed as:
# (sqrt2/2).sin ((5pi)/12).#
We can evaluate #sin ((5pi)/12)# by using the trig identity:
#cos 2a = 1 - 2sin^2 a#
#cos ((10pi)/12) = cos ((5pi)/6) = -sqrt3/2 = 1 - 2sin^2 ((5pi)/12)#
#2sin^2 ((5pi)/12 = 1 + sqrt3/2 = (2 + sqrt3)/2#
#sin^2 ((5pi)/12) = (2 + sqrt3)/4#
#sin ((5pi)/12) = sqrt(2 + sqrt3)/2# (note: sin ((5pi)/12) is positive)
Finally,
#P = (sqrt2/4)(sqrt(2 + sqrt3))#