How do you use the half-angle identity to find the exact value of csc [ (7pi) / 8 ]?

2 Answers
Vinícius Ferraz
Aug 20, 2015

#2 / sqrt {2 - sqrt 2}#

Explanation:

#n = cossec \frac{7 pi}{8} = \frac{1}{sin \frac{7 pi}{8}} = \frac{1}{sin \frac{pi}{8}}#

#cos 2x = cos^2 x - sin^2 x = 1 - 2 sin^2 x#

#cos frac{pi}{4} = 1 - 2 sin^2 frac{pi}{8}#

#sqrt 2 /2 = 1 -2 frac{1}{n^2}#

#frac{2}{n^2} = 1 - sqrt 2 / 2#

#frac{2}{n^2} = (2 - sqrt 2) / 2#

#n^2 = 4 / (2 - sqrt 2)#

#n = 2 / sqrt {2 - sqrt 2}#

Nghi N.
Aug 20, 2015

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

Ans: #2/(sqrt(2 - sqrt2)#

Explanation:

#csc ((7pi)/8) = 1 /sin ((7pi)/8)#
Find #sin ((7pi)/8)#. Call #sin ((7pi)/8) = sin t#
#cos 2t = cos ((14pi)/8) = cos ((7pi)/4) = cos (pi/4) = sqrt2/2#
Apply the trig identity: #cos 2t = 1 - 2sin^2 t #

#cos 2t = sqrt2/2 = 1 - 2sin^2 t#
#2sin^2 t = 1 - sqrt2/2 = (2 - sqrt2)/2#
#sin^2 t = (2 - sqrt2)/4 #
#sin ((7pi)/8) = sin t = +- [sqrt(2 - sqrt2)]/2#
Since the arc (7pi/8) is located in Quadrant II. then only the positive answer is accepted.
#sin ((7pi)/8) = (sqrt(2 - sqrt2))/2#
Therefor, #csc ((7pi)/8) = (2)/(sqrt(2 - sqrt2)#

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