How do you use a half-angle formula to simplify #csc [ (7pi) / 8 ]#?

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Answer 1 · 2015-07-06T04:56:07

Find #csc ((7pi)/8)#

#csc ((7pi)/8) = 1/sin ((7pi)/8)#

Call #sin ((7pi)/8) = sin a# (Quadrant II)

#cos 2a = cos ((14pi)/8) = cos ((pi)/4 + 2pi) = cos ((pi)/4) = (sqrt2)/2#

Use trig identity #cos 2a = 1 - 2sin^2 a.#

#(sqrt2)/2 = 1 - 2sin^2a# --> #sin^2 a = (2 - sqrt2)/4#

#sin a = sin ((7pi)/8) = sqrt(2 + sqrt2)/2#

#csc ((7pi)/8) = 1/sin a = 2/(sqrt(2 + sqrt2)#