Solving Trigonometric Equation with two angles I have the following equation, trying to solve for y:
$$A\sin(y)=(B-A\cos(y))\tan(x)$$
I got it to:
$$(A\sin(y))/B - \tan(y) = \tan(x)$$
But don't really see anything to do from there. Any help is appreciated.
 A: simplifying your term you will get
$$A\sin(y)+A\tan(x)\cos(y)=B\tan(x)$$ and now use that
$$\sin(y)=2\,{\frac {\tan \left( y/2 \right) }{1+ \left( \tan \left( y/2
 \right)  \right) ^{2}}}
$$
and $$\cos(y)={\frac {1- \left( \tan \left( y/2 \right)  \right) ^{2}}{1+ \left( 
\tan \left( y/2 \right)  \right) ^{2}}}
$$ and set $$\tan(y/2)=t$$ and solve your equation for $t$
A: Hint
$$A\sin(y)=(B-A\cos(y))\cdot\tan(x)$$ Multiply by $\cos(x)$ $$A \sin(y)\cos(x)=B \tan(x)\cos(x)-A\cos(y)\tan(x)\cos(x)$$ $$A \sin(y)\cos(x)=B\sin(x)-A\cos(y)\sin(x)$$ $$A\sin(x+y)=B\sin(x)$$ I am sure that you can take it from here.
A: Hint
$$A\sin(y)=(B-A\cos(y))\cdot\tan(x)=A[\sin (y)+\tan (x)\cdot \cos (y)]=B\tan (x)$$
If you just work with $\sin (y)+\tan (x)\cdot \cos (y)$ you can use
$$\sqrt{1+\tan^2(x)}\left(\frac{1}{\sqrt{1+\tan^2(x)}}\cdot \sin (y)+\frac{\tan (x)}{\sqrt{1+\tan^2(x)}}\cdot \cos (y)\right)=\sqrt{1+\tan^2(x)}\left(\cos \theta\cdot \sin (y)+\sin \theta\cdot \cos (y)\right)=\sqrt{1+\tan^2(x)}\left(\sin(y+\theta)\right)$$
