How do you find #cos67.5# using the half-angle identity?

2 Answers
Ratnaker Mehta
Jul 25, 2016

#cos(67.5^@)=+(sqrt(2-sqrt2))/2#

Explanation:

The Half-Angle Identity is #1+costheta=2cos^2(theta/2)#

Taking, #theta=2xx67.5^@=135^@, 1+cos(135^@)=2cos^2(67.5^@)#

#:. 2cos^2(67.5^@)=1+cos(180^@-45^@)=1-cos45^@#

#=1-1/sqrt2=(sqrt2-1)/sqrt2=(sqrt2-1)/sqrt2*sqrt2/sqrt2=(2-sqrt2)/2#

#:.cos^2(67.5^@)=(2-sqrt2)/4#

#:. cos(67.5^@)=+(sqrt(2-sqrt2))/2#, #+ve# sign, as, 67.5^@ is in the first Quadrant#

Truong-Son N.
Jul 25, 2016

I got #sqrt(2 - sqrt2)/2# as well.

Since #67.5^@# is half of #135^@#, let's work off of that. I remember it by starting from the #cos^2(x)# identity.

#cos^2(x) = (1 + cos(2x))/2#

#cos^2(x/2) = (1 + cosx)/2#

#:. cos(x/2) = pmsqrt((1 + cosx)/2)#

Since #67.5^@# is in quadrant #"I"#, where #0 < x < 90^@#, and #cos(x)# for #0 < x < 90^@# is positive, #cos(67.5^@) > 0#. Therefore:

#cos(135^@/2) = +sqrt((1 + cos(135^@))/2)#

#= sqrt((1 + cos((3pi)/4))/2)#

#= sqrt((1 - sqrt2/2)/2)#

#= sqrt((2/2 - sqrt2/2)/2) = sqrt(((2 - sqrt2)/2)/2)#

#= sqrt((2 - sqrt2)/4)#

#= color(blue)(sqrt(2 - sqrt2)/2 ~~ 0.3827)#

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