Question

Express 1/(cos θ − i sin θ) in the form of a + ib and hence prove that

cosθ+isinθ/ cos θ − i sinθ = cos2θ+isin2θ.

Answer 1

$$\dfrac{1}{\cos \theta − i \sin \theta}=\dfrac{1}{\cos \theta − i \sin \theta}\cdot\dfrac{\cos \theta + i \sin \theta}{\cos \theta + i \sin \theta}$$

$$=\dfrac{\cos \theta + i \sin \theta}{\cos^2 \theta + \sin^2 \theta}=\dfrac{\cos \theta + i \sin \theta}{1}$$

$$=\cos \theta + i \sin \theta$$

$$\dfrac{\cos \theta + i \sin \theta}{\cos \theta - i \sin \theta}=(\cos \theta + i \sin \theta)(\cos \theta + i \sin \theta)$$

$$=\cos^2 \theta+i\cos \theta\sin \theta+i\sin \theta\cos \theta-\sin^2 \theta$$

$$=(\cos^2 \theta-\sin^2 \theta)+i(2\cos \theta\sin \theta)$$

$$=\cos(2\theta)+i\sin(2\theta)$$