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

1 Answer
Jul 11, 2016

#P = (1/4)sqrt(2 - sqrt2)#

Explanation:

#P = cos (pi/3).sin (pi/8)#
Trig table --> cos (pi/3) = 1/2.
There for, P can be expressed as #P = (1/2)sin (pi/8)#
We can evaluate sin (pi/8) by using trig identity:
cos 2a = 1 - 2sin^2 a
#cos (2pi)/8 = cos (pi/4) = sqrt2/2 = 1 - 2sin^2 (pi/8)#
#2sin^2 (pi/8) = 1 - sqrt2/2 = (2 - sqrt2)/2#
#sin^2 (pi/8) = (2 - sqrt2)/4#
#sin (pi/8) = +- sqrt(2 - sqrt2)/2#
Since #sin (pi/8)# is positive, then,
#P = (1/4)sqrt(2 - sqrt2)#