How do you use half-angle formulas to find exact values of #sin((-17pi)/12)# and #cos((-17pi)/12)#?

1 Answer
Nghi N.
Jun 9, 2015

Call #t = (-17pi)/12#
Use the trig identity: #2sin^2 t = 1 - cos 2t #

Explanation:

#sin t = sin (2pi - (17pi)/12) = sin ((7pi)/12)# =

#cos ((14pi)/12) = cos ((2pi)/12 + pi) = cos ((2pi)/12) = cos (pi/6) = 1/2#

#2sin^2 t = 1 - cos 2t = 1 - 1/2 = 1/2#

#sin^2 t = 1/4# --> #sin t = sin ((-17pi)/12) = +- 1/2#

#2cos^2 t = cos 2t + 1 = 1/2 + 1 = 3/2#

#cos^2 t = 3/4 -> cos t = cos ((-17pi)/12) = +- sqrt3/2#

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