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:

$cos^2x=cos^2x$

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