How do you use the half angle identity to find exact value of sin(-15)?

1 Answer
Alan P.
Sep 6, 2015

#sin(-15^@) = -sqrt(2-sqrt(3))/2 ~~ -0.25882#

Explanation:

Half angle formula for #sin(theta)#
#color(white)("XXXX")sin(theta/2) = +-sqrt((1-cos(theta))/2)#

#color(white)("XXXX")sin(-15^@) = -sqrt((1-cos(-30)/2)#
#color(white)("XXXXXXXX")#...we can rule out the positive version as extraneous since #sin# is negative in Q IV

#cos(-30^@) = sqrt(3)/2#
enter image source here

#sin(-15^@) = -sqrt((1-sqrt(3)/2)/2) = - sqrt(2-sqrt(3))/2#

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