How do you show that $cos(2x) = 2cos^2x- 1$ ?

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Answer 1 · 2016-11-10T16:32:19

$cos(2x) = cos(x + x)$

By the sum and difference identities, we have that $cos(A + B) = cosAcosB - sinAsinB$ .

$cos(x+ x) = cosxcosx - sinxsinx$

$cos(2x) = cos^2x - sin^2x$

Use the identity $sin^2theta + cos^2theta = 1 -> sin^2theta = 1 - cos^2theta$ .

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

$cos(2x) = cos^2x - 1 + cos^2x$

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

$LHS = RHS$

This identity has been proved!

Hopefully this helps!

How do you show that cos(2x) = 2cos^2x- 1cos(2x) = 2cos^2x- 1?