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I am working on a program which will predict the tides, but have come across a problem when using the simplified harmonic method of tidal prediction, I understand the whole thing but cannot do the following, this is what I have so far:

$R\sin(r) = H\sin(\theta)$

$R\cos(r) = H\cos(\theta)$

How do I obtain the values of just R and r alone?

EDIT 2! it is slightly more complex than first explained, this is more like what I am trying to work with:

$R\sin(r) = A\sin(Y) + B\sin(Z)$

$R\cos(r) = A\cos(Y) + B\cos(Z)$

Thanks.

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    Final question update! The angles were not the same, there are two angles used, Y and Z.2012-10-22

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You know that $\sin^2\theta+\cos^2\theta=1$. Hence, by squaring both equations and adding you have: $R^2=R^2(\sin^2r+\cos^2r)=H^2(\sin^2\theta+\cos^2\theta)=H^2$ Hence $R=\pm H$. From here you know that $\sin r=\pm\sin\theta$ (depending on the value of $H$). Assuming that $H,R\geq0$ then you have $\sin r=\sin\theta$. So $r=\theta+2\pi k$

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    @AlexanderGruber I know this is a long time ago, but I used your formula for the application and it worked but now I'm revisiting it as I didn't really understand why it worked. Would you be able to explain? When I try to work it out I only get $R^2cos^2(r) + R^2sin^2(r)$ simplifies to $R^2$ and the same with the rest leaving $R^2 = A^2 + B^2$ where does the $+2ABcos(y-z)$ come from? Thanks2015-06-24