Question

if x=cis alpha, y=cis beta, then prove that(x+y)(xy-1) /(x-y) (xy+1)=sin alpha+sin beta/sinalpha-sin beta
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Answer 1

$\textbf{Given:}$

$\mathsf{x=cos\alpha+i\,sin\alpha}$

$\mathsf{y=cos\beta+i\,sin\beta}$

$\textbf{To prove:}$

$\mathsf{\dfrac{(x+y)(xy-1)}{(x-y)(xy+1)}=\dfrac{sin\alpha+sin\beta}{sin\alpha-sin\beta}}$

$\textbf{Solution:}$

$\mathsf{Consider,}$

$\mathsf{x=cos\alpha+i\,sin\alpha}$

$\mathsf{y=cos\beta+i\,sin\beta}$

$\mathsf{Then,}$

$\mathsf{\bar{x}=cos\alpha-i\,sin\alpha}$

$\mathsf{\bar{y}=cos\beta-i\,sin\beta}$

$\implies\mathsf{\dfrac{x-\bar{x}}{2i}=sin\alpha\;\;\&\;\;\dfrac{y-\bar{y}}{2i}=sin\beta}$

$\mathsf{Now}$

$\mathsf{\dfrac{sin\alpha+sin\beta}{sin\alpha-sin\beta}}$

$\mathsf{=\dfrac{\left(\dfrac{x-\bar{x}}{2i}\right)+\left(\dfrac{y-\bar{y}}{2i}\right)}{\left(\dfrac{x-\bar{x}}{2i}\right)-\left(\dfrac{y-\bar{y}}{2i}\right)}}$

$\mathsf{=\dfrac{\dfrac{(x-\bar{x})+(y-\bar{y})}{2i}}{\dfrac{(x-\bar{x})-(y-\bar{y})}{2i}}}$

$\mathsf{=\dfrac{(x-\bar{x})+(y-\bar{y})}{(x-\bar{x})-(y-\bar{y})}}$

$\mathsf{=\dfrac{(x+y)-(\bar{x}+\bar{y})}{(x-y)-(\bar{x}-\bar{y})}}$

$\textsf{We know that,}$

$\boxed{\mathsf{If\;|z|=1,\;then\;\bar{z}=\dfrac{1}{z}}}$

$\mathsf{=\dfrac{(x+y)-(\dfrac{1}{x}+\dfrac{1}{y})}{(x-y)-(\dfrac{1}{x}-\dfrac{1}{y})}}$

$\mathsf{=\dfrac{(x+y)-(\dfrac{x+y}{xy})}{(x-y)-(\dfrac{y-x}{xy})}}$

$\mathsf{=\dfrac{(x+y)-(\dfrac{x+y}{xy})}{(x-y)+(\dfrac{x-y}{xy})}}$

$\mathsf{=\dfrac{(x+y)(1-\dfrac{1}{xy})}{(x-y)(1+\dfrac{1}{xy})}}$

$\mathsf{=\dfrac{(x+y)(\dfrac{xy-1}{xy})}{(x-y)(\dfrac{xy+1}{xy})}}$

$\mathsf{=\dfrac{(x+y)(xy-1)}{(x-y)(xy+1)}}$

$\implies\boxed{\mathsf{\dfrac{(x+y)(xy-1)}{(x-y)(xy+1)}=\dfrac{sin\alpha+sin\beta}{sin\alpha-sin\beta}}}$

$\textbf{Find more:}}$

Find z if arg(z+2i)=π/4 and arg(z-2i)=3π/4 solutions

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