How do you express #sin(pi/ 4 ) * sin( ( 7 pi) / 8 ) # without using products of trigonometric functions?

1 Answer
Nghi N.
Apr 7, 2016

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

Explanation:

#P = sin (pi/4).sin ((7pi)/8)#
Trig table --> sin (pi/8) = sqrt2/2
Trig unit circle -->
#sin ((7pi)/8) = sin (pi/8)#
Evaluate sin (pi/8) by using the thig identity:
#cos 2a = 1 - 2sin^2 a#
#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# --> #sin (pi/8)# is positive.
Finally:
#P = (sqrt2)((sqrt(2 - sqrt2)]/4)#

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