The A + iB form of frac{(cos x + isin x)(cos | vedclass

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Question

(A) Given expression: $Z = \frac{(\cos x + i\sin x)(\cos y + i\sin y)}{(\cot u + i)(1 + i	an v)}$Using Euler's formula,the numerator is $e^{i(x+y)}$.

Vedclass · Accepted Answer

$\sin u \cos v [\cos (x + y - u - v) + i\sin (x + y - u - v)]$

Vedclass · Answer

(A) Given expression: $Z = \frac{(\cos x + i\sin x)(\cos y + i\sin y)}{(\cot u + i)(1 + i	an v)}$Using Euler's formula,the numerator is $e^{i(x+y)}$.For the denominator:$(\cot u + i) = \frac{\cos u + i\sin u}{\sin u} = \frac{e^{iu}}{\sin u}$$(1 + i	an v) = \frac{\cos v + i\sin v}{\cos v} = \frac{e^{iv}}{\cos v}$Substituting these into the expression:$Z = \frac{e^{i(x+y)}}{\frac{e^{iu}}{\sin u} \cdot \frac{e^{iv}}{\cos v}} = \sin u \cos v \cdot \frac{e^{i(x+y)}}{e^{i(u+v)}}$$Z = \sin u \cos v \cdot e^{i(x+y-u-v)}$$Z = \sin u \cos v [\cos(x+y-u-v) + i\sin(x+y-u-v)]$

The $A + iB$ form of $\frac{(\cos x + i\sin x)(\cos y + i\sin y)}{(\cot u + i)(1 + i\tan v)}$ is

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