Answer to Question #282432 in Complex Analysis for iqa

Question #282432

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.


1
Expert's answer
2021-12-28T13:26:30-0500
1cosθisinθ=1cosθisinθcosθ+isinθcosθ+isinθ\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}

=cosθ+isinθcos2θ+sin2θ=cosθ+isinθ1=\dfrac{\cos \theta + i \sin \theta}{\cos^2 \theta + \sin^2 \theta}=\dfrac{\cos \theta + i \sin \theta}{1}

=cosθ+isinθ=\cos \theta + i \sin \theta

cosθ+isinθcosθisinθ=(cosθ+isinθ)(cosθ+isinθ)\dfrac{\cos \theta + i \sin \theta}{\cos \theta - i \sin \theta}=(\cos \theta + i \sin \theta)(\cos \theta + i \sin \theta)

=cos2θ+icosθsinθ+isinθcosθsin2θ=\cos^2 \theta+i\cos \theta\sin \theta+i\sin \theta\cos \theta-\sin^2 \theta

=(cos2θsin2θ)+i(2cosθsinθ)=(\cos^2 \theta-\sin^2 \theta)+i(2\cos \theta\sin \theta)

=cos(2θ)+isin(2θ)=\cos(2\theta)+i\sin(2\theta)


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