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

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Answer 1 · 2017-01-06T03:39:46

#cos ((7pi)/8)= -sqrt ((sqrt2 +1)/(2sqrt2))#

Write cos#(7pi)/8# = cos #(pi-pi/8)#= -cos #pi/8#

Now using half angle formula, cosx= #2 cos^2 (x/2) -1#, we can write cos #pi/4# = 2 #cos^2 (pi/8) -1#

This means #1/sqrt2 +1 = 2 cos^2 (pi/8)#

#2cos^2 (pi/8) = (sqrt2 +1)/sqrt2#

#cos^2 (pi/8) = (sqrt2 +1)/(2sqrt2)#

#cos (pi/8)= sqrt ((sqrt2 +1)/(2sqrt2))#

Hence #cos ((7pi)/8)= - cos (pi/8) = -sqrt ((sqrt2 +1)/(2sqrt2))#