How do you prove #cos^2x - sin^2x = 2cos^2x - 1#?

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Answer 1 · 2015-06-13T19:58:58

Have a look:

Given:

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

we can write it as (taking #-1# to the left and #cos^2x# to the right):

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

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

But #sin^2x+cos^2x=1#;
then: #1-sin^2x=cos^2x#;
so:

#cos^2x=cos^2x#