How do you find the value of #sin((7pi)/8)# using the double or half angle formula?

1 Answer
Nghi N.
Sep 18, 2016

#sqrt(2 - sqrt2)/2#

Explanation:

Trig table of special arcs, and unit circle -->
#sin ((7pi/8) = sin (pi - pi/8) = sin (pi/8)#
Find #sin (pi/8)# by using trig identity:
#2sin^2 a = 1 - cos 2a#
#2sin^2 (pi/8) = 1 - cos (pi/4) = 1 - sqrt2/2 = (2 - sqrt2)/2#
#sin^2 (pi/8) = (2 - sqrt2)/4#
#sin (pi/8) = sqrt(2 - sqrt2)/2# (because #sin (pi/8)# is positive)
Finally,
#sin ((7pi)/8) = sin (pi/8) = sqrt(2 -sqrt2)/2#

Related questions
See all questions in Half-Angle Identities
Impact of this question
8048 views around the world
You can reuse this answer
Creative Commons License