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)?

Aug 25, 2015

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

Explanation:

$\cos \left(2 x\right) + i \sin \left(2 x\right)$

$= {e}^{2 i x} = {\left({e}^{i x}\right)}^{2}$

$= {\left(\cos \left(x\right) + i \sin \left(x\right)\right)}^{2}$

$= {\cos}^{2} \left(x\right) + 2 i \sin \left(x\right) \cos \left(x\right) + {i}^{2} {\sin}^{2} \left(x\right)$

$= \left({\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right) + i \left(2 \sin \left(x\right) \cos \left(x\right)\right)$

Equate real and imaginary parts to get:

$\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$

$\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$