How do you verify the identity: #cos^2x-sin^2x=1-2sin^2x#?

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Alan P. Share
Jun 2, 2017

From the basic definitions of #sin# and #cos# and the Pythagorean Theorem
#cos^2(x)+sin^2(x) = 1#

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

So
#cos^2(x) - sin^2(x)#

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

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

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Write your answer here...
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Jim H Share
Apr 22, 2015

There are many acceptable (correct) ways to do this.

Here's one:

(use #sin^2x+cos^2x = 1#

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

#color(white)"ssssssssssss"# #=cos^2x+sin^2x-sin^2x-sin^2x#

#color(white)"ssssssssssss"# #=cos^2x-sin^2x#

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Tom Share
Apr 22, 2015

#cos^2(x)-sin^2(x)#

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

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