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I met some questions about the following system of equations:

$$f=\frac{\sin[a\cdot\sin(\alpha)\cdot\cos(\beta)]}{\sin[b\cdot\sin(\alpha)\cdot\cos(\beta)]} \times \frac{\sin[a\cdot\sin(\alpha)\cdot\sin(\beta)]}{\sin[b\cdot\sin(\alpha)\cdot\sin(\beta)]} $$

Where a,b and $\beta$ are constants.

First of all, I want to calculate the derivative of the function $f$ with Wolfram Mathematica 10.4 :
$$y=\frac{df}{d\alpha}$$ But the final outcome is very complicated.I try to use "simplify" and "fullsimplify" commands to simplify the final outcome but failed.

Then,I want to know when $\alpha$ is choosen how much, can make the function $y=0$. certainly,I have do this by Wolfram Mathematica 10.4,but it is failed.The reason is that it is failed in the previous step and function $y$ is vary complicated.

Thank you very much! I am very upset,because I have to do it for a long time.

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To simplify your expression $$f=\frac{\sin[a\cdot\sin(\alpha)\cdot\cos(\beta)]}{\sin[b\cdot\sin(\alpha)\cdot\cos(\beta)]} \times \frac{\sin[a\cdot\sin(\alpha)\cdot\sin(\beta)]}{\sin[b\cdot\sin(\alpha)\cdot\sin(\beta)]}$$ use more compact notations since you are considering $\beta$ as a constant $$f=\frac{\sin[A\sin(\alpha)]}{\sin[C\sin(\alpha)]} \times \frac{\sin[B\sin(\alpha)]}{\sin[D\sin(\alpha)]}$$ and use product and quotient rule (or logarithmic differentiation) remembering that $$\frac d {d\alpha}\sin[k \sin(\alpha)]=k \cos (\alpha ) \cos [k \sin (\alpha )]$$ You should end with a simpler "monster" for $\frac{df}{d\alpha}$.

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  • $\begingroup$ Thank you very much! I will try it! with logarithmic differentiation method. Thank you very mach for your kind explanation! $\endgroup$ – 张兴民 Feb 24 '17 at 9:02
  • $\begingroup$ @张兴民. You are very welcome ! $\endgroup$ – Claude Leibovici Feb 24 '17 at 9:04
  • $\begingroup$ .Yes,the end result is very long and like simpler "monster" .I try to sinplify the "monster", because next I want to solve the function. But now I I don't have any ideals to sinplify the "monster". $\endgroup$ – 张兴民 Feb 24 '17 at 9:29
  • $\begingroup$ @张兴民. The monster fits in 2 lines and it is ugly ! I have no hope for analytical solutions beside $0$ and $\frac \pi 2$. $\endgroup$ – Claude Leibovici Feb 24 '17 at 9:47

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