If #cos^2x = 3/4#, then what is the value of #cos(2x)#?

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Answer 1 · 2017-05-29T02:43:46

#cos(2x) = 1/2#

Use the identity #cos^2(x)+sin^2(x) = 1# to find the value of #sin^2(x)#:

#3/4 + sin^2(x) = 1#

#sin^2(x) = 1/4#

Use the identity #cos(2x) = cos^2(x) - sin^2(x)#

#cos(2x) = 3/4 - 1/4#

#cos(2x) = 1/2#

Answer 2 · 2017-05-29T03:55:21

#cos(2x) = 1/2#

We have

#cos(2x) = 2cos^2x- 1#

#cos(2x) = 2(3/4) - 1#

#cos(2x) = 3/2 - 1#

#cos(2x) = 1/2#

Hopefully this helps!