Question 1187233: Could you please help me prove that 2cos(x)^2 - 1 = cos(2x)?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! here's a summary of trig identities that can help.
https://www2.clarku.edu/faculty/djoyce/trig/identities.html
the trig identity you want is:
cos(2*theta) = cos^2(theta) - sin^2(theta)
the other identity that you want is:
sin^2(theta) + cos^2(theta) = 1
in this other identity, solve for sin^2(theta) to get:
sin^2(theta) = 1 - cos^2(theta).
in the first identity of cos(2*theta) = cos^2(theta) - sin^2(theta), replace sin^2(theta) with 1 - cos^2(theta) to get:
cos(2*theta) = cos^2(theta) - (1 - cos^2(theta)).
simplify to get:
cos(2*theta) = cos^2(theta - 1 + cos^2(theta)
combine like terms to get:
cos(2*theta) = 2cos^2(theta) - 1
note that 2cos^2(theta) is the same as 2cos(theta)^2.
you get:
cos(2*theta) = 2cos(theta)^2 - 1
that proves that 2cos(x)^2 - 1 = cos(2x)
trust me that cos^2(x) is equal to cos(x)^2.
if you tried to do cos^2(x) in your calculator, it wouldn't be able to do it.
you would have to enter cos^2(x) as cos(x)^2.
at least that's the way it is with my ti-84 plus calculator.
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