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I'm trying to understand how to simplify this equation:

$$\beta =\sin { \left( 2\arctan { \frac { y }{ x } -\frac { \pi }{ 2 } } \right) } $$

I know the result is $β = (y²-x²)∕(y²+x²)$ but how can I simplify this step by step?

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    1) Use $\sin(\alpha+\beta)$ formula to expand the sum inside the sine. 2) Use [this](https://wikimedia.org/api/rest_v1/media/math/render/svg/49a096f6b0f0b866d9286e274ce5404e52aeb902) formula with $\theta = \arctan(y/x)$. – 2017-02-15

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Hint. Some useful steps. $$ \begin{align} &\sin\left(a-\frac \pi2\right)=-\sin\left(\frac \pi2-a\right)=-\cos a \\\\ &\cos 2a=\frac{1-\tan^2 a}{1+\tan^2 a} \end{align} $$ Then just apply it with $\displaystyle a=\arctan \frac yx$.

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    Thanks a lot, sorry to ask but is there more details about how to apply arctany/x to it? – 2017-02-15
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    @ToddLiang Sure. Then you have $$\beta =\sin \left( 2\arctan \frac yx -\frac {\pi }{ 2 } \right) =-\cos\left( 2\arctan \frac { y }{ x } \right)=-\frac{1-\tan^2 (\arctan y/x)}{1+\tan^2(\arctan y/x)} =-\frac{1-y^2/x^2}{1+y^2/x^2}$$ that is $\beta= -\frac{1-y^2/x^2}{1+y^2/x^2} =\frac{y^2-x^2}{y^2+x^2}$ as announced. We have used that $\tan(\arctan y/x)=y/x$. – 2017-02-15
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Hint: Use $\sin(x-\dfrac{\pi}{2})=-\cos x$ and then $$\cos2\alpha=\frac{1-\tan^2\alpha}{1+\tan^2\alpha}$$