suppose I have an equations of the following with two unknowns $A$ and $\theta$


I have two points $(E,F) (G,H)$ how do I go about solving this equation analytically. I can solve this equation by using least squares where I just plugin a few numbers and solve it iteratively.

I was thinking about using trig identities and breaking it down to $A\sin(x)\cos(\theta)+Acos(x)\sin(\theta)=D$

But I am kind of stuck at that point. Using derivative to solve the equation doesn't help since the form of the equation is still $\cos(x+\theta)$

  • 1
    $\begingroup$ Does this answer your question? $\endgroup$ Commented Sep 14, 2012 at 18:00
  • $\begingroup$ The A term in my equation is still inside the arcsin function. I am not sure how to get it out of the arcsin function. $\endgroup$ Commented Sep 14, 2012 at 18:27
  • $\begingroup$ I'm sorry, I thought you wanted to solve for $x$. Where do you use $(E,F)$ and $(G,H)$ by the way? $\endgroup$ Commented Sep 14, 2012 at 18:33
  • $\begingroup$ Here would be the form of my two equations: Asin(E+θ)=F,Asin(G+θ)=H. I am trying to solve for A, θ $\endgroup$ Commented Sep 14, 2012 at 18:38
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    $\begingroup$ OK, now I understand your problem. Your idea of splitting the $\sin$ is a good one. Call $x=A\cos(\theta)$ and $y=A\sin(\theta)$. You then have two linear equations in $x$ and $y$. Solve these. Then, compute $x^2+y^2$ and $y/x$. Do you see how to work from there? $\endgroup$ Commented Sep 14, 2012 at 18:47

1 Answer 1


I give all credit to Raskolnikov

Given this equations: $A\sin(x+θ)=D$

and these two points: $(E,F),(G,H)$

So these are the set of equations: $A\sin(E+θ)=F, A\sin(G+θ)=H$

so first expand them: $A\sin(E)\cos(θ)+A\cos(E)\sin(θ)=F, A\sin(G)\cos(θ)+A\cos(G)\sin(θ)=H$

Then let $X=A\cos(θ), Y=A\sin(θ)$

so your equations are: $X\sin(E)+Y\cos(E)=F, X\sin(G)+Y\cos(G)=H$

now solve for $X$ and $Y$.

Now set: $A=X/\cos(θ)$

So that: $Y=X\sin(θ)/\cos(θ) = Y/X=\tan(θ)$

You solve for $\theta$,

You can substitute $\theta$ into one of the previous equation and solve for A.


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