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

Question

12749 views around the world

You can reuse this answer
Creative Commons License

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!