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Prove that $$\frac{\sin(x+\theta)}{\sin(x+\phi)} = \cos(\theta - \phi) +\cot(x+\phi)\sin(\theta-\phi).$$

I tried applying componendo and dividendo on LHS but couldn't prove. Please help.

Thank you.

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  • $\begingroup$ Use $\cot=\frac{\cos}{\sin}$ in right hand and simplified it $\endgroup$ – Nosrati Jan 27 '17 at 16:44
  • $\begingroup$ @MyGlasses thanks i proves lhs from rhs. but how to proceed from compo and dividendo is my concern here $\endgroup$ – Sophie Clad Jan 27 '17 at 16:50
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    $\begingroup$ Can you share how you attempted compo and dividendo? $\endgroup$ – imranfat Jan 27 '17 at 16:51
  • $\begingroup$ @imranfat just after applying it and used sinx+siny formula $\endgroup$ – Sophie Clad Jan 27 '17 at 17:06
  • $\begingroup$ @SophieClad Instead of prove one problem twice, prove two problems. $\endgroup$ – Nosrati Jan 27 '17 at 17:23
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$$\frac{\sin(x+\theta)}{\sin(x+\phi)}=\frac{\sin[(x+\phi)+(\theta-\phi)]}{\sin(x+\phi)}=\frac{\sin(x+\phi)\cos(\theta-\phi)+\sin(\theta-\phi)\cos(x+\phi)}{\sin(x+\phi)}=\\ =\frac{\sin(x+\phi)\cos(\theta-\phi)}{\sin(x+\phi)}+\frac{\sin(\theta-\phi)\cos(x+\phi)}{\sin(x+\phi)}=\cos(\theta-\phi)+\cot(x+\phi)\sin(\theta-\phi)$$

for the last equality I used $\cot(x+\phi)=\frac{\cos(x+\phi)}{\sin (x+\phi)}$

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Multiply both sides by $\sin(x+ϕ)$: $$\sin(x+θ) = \sin(x+ϕ)\cos(θ−ϕ)+\cos(x+ϕ)\sin(θ−ϕ) = \sin(x+ϕ+θ−ϕ)$$ (by the angle sum formula, and the fact that $\cot x=\frac{\cos x}{\sin x}$)

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We have that $$\begin{align}\frac{\sin(x+\theta)}{\sin(x+\phi)}&=\frac{\sin\left((x\color{red}{+\phi})+(\theta\color{red}{-\phi})\right)}{\sin(x+\phi)} \\ &=\frac{\sin(x+\phi)\cos(\theta-\phi)+\sin(\theta-\phi)\cos(x+\phi)}{\sin(x+\phi)}& \\ \\ &=\frac{\color{blue}{\sin(x+\phi)}\cos(\theta-\phi)}{\color{blue}{\sin(x+\phi)}}+\frac{\sin(\theta(\phi)\color{green}{\cos(x+\phi)}}{\color{green}{\sin(x+\phi)}} \\ &=\cos(\theta-\phi)+\color{green}{\cot(x+\phi)}\sin(\theta-\phi)\end{align}$$

by the angle sum formula.

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