Question

Express 1/(cosθ − i sinθ) in the form of a + ib and hence prove that

cos θ + i sinθ

cos θ − i sinθ = cos 2θ + i sin2θ.

Answer 1

$\frac{1}{cos\theta -isin\theta}=\frac{cos\theta +isin\theta}{(cos\theta -isin\theta)(cos\theta +isin\theta)}=\frac{cos\theta +isin\theta}{cos^2\theta +sin^2\theta}=cos\theta +isin\theta$

$\frac{cos\theta +isin\theta}{cos\theta -isin\theta}=(cos\theta +isin\theta)^2=cos^2\theta +2isin\theta cos\theta-sin^2\theta= cos 2θ + i sin2θ$