How do you prove #cos(2x) = [cos(x)]^2 - [sin(x)]^2# and #sin(2x) = 2sin(x)cos(x)# using Euler's Formula: #e^(ix) = cos(x) + isin(x)#?

Question

2273 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-25T06:47:14

Use Euler's Formula to evaluate #cos(2x)+i sin(2x)#, then equate real and imaginary parts to get the double angle formulae.

#cos(2x)+i sin(2x)#

#= e^(2ix) = (e^(ix))^2#

#= (cos(x) + i sin(x))^2#

#= cos^2(x)+2i sin(x)cos(x) + i^2sin^2(x)#

#= (cos^2(x) - sin^2(x)) + i(2sin(x)cos(x))#

Equate real and imaginary parts to get:

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

#sin(2x) = 2 sin(x)cos(x)#