How do I solve the following trigonometric equations for $\alpha$ and $\beta$:

$$ x = d\cos(\alpha + \beta)+h\cos(\alpha) $$ $$ y = d\sin(\alpha + \beta)+h\sin(\alpha) $$

My attempt:

$$ \sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$$ $$ \cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$$


$$ x = d\cos(\alpha + \beta)+h\cos(\alpha) \implies d(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta))+h\cos(\alpha)=x$$ $$ y = d\sin(\alpha + \beta)+h\sin(\alpha) \implies d(\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta))+h\sin(\alpha) = y$$

Solve the first equation for $\beta$:

$$\frac{x-h\cos(\alpha)}{d} = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \implies ??\text{ (I'm stuck here) }$$

I would be very grateful for any hints about how to continue from there.

  • $\begingroup$ If we look at a double pendulum for a while, we'll see that $\sqrt{x^2+y^2}$ must depend solely on $\beta$ and not on $\alpha$. Thus we may find $\beta$, and then go back and plug it into something to find $\alpha$ as well. $\endgroup$ – Ivan Neretin Feb 19 '16 at 22:10
  • $\begingroup$ Near the the fixed point length and angle are $h,\alpha$ ; At mid point $ d, (\alpha+\beta)$ $\endgroup$ – Narasimham Feb 19 '16 at 22:30

After squaring you have \begin{align} x^2&=d^2\cos^2(\alpha+\beta)+2dh\cos(\alpha+\beta)\cos\alpha+h^2\cos^2\alpha\\ y^2&=d^2\sin^2(\alpha+\beta)+2dh\sin(\alpha+\beta)\sin\alpha+h^2\sin^2\alpha \end{align} Summing them up gives $$ 2dh(\cos(\alpha+\beta)\cos\alpha+\sin(\alpha+\beta)\sin\alpha) =x^2+y^2-d^2-h^2 $$ The left hand side is clearly $\cos\beta$, so we have got $$ \cos\beta=\frac{x^2+y^2-d^2-h^2}{2dh} $$ Expanding the equations with the addition formulas gives $$ \begin{cases} (d\cos\beta+h)\cos\alpha-d\sin\beta\sin\alpha=x \\[6px] d\sin\beta\cos\alpha+(d\cos\beta+h)\sin\alpha=y \end{cases} $$ and Cramer's rule provides $$ \begin{cases} \cos\alpha= \dfrac {x(d\cos\beta+h)+dy\sin\beta} {(d\cos\beta+h)^2+d^2\sin^2\beta} = \dfrac {x(d\cos\beta+h)+dy\sin\beta} {x^2+y^2} \\[8px] \sin\alpha= \dfrac {y(d\cos\beta+h)+dx\sin\beta} {(d\cos\beta+h)^2+d^2\sin^2\beta} = \dfrac {y(d\cos\beta+h)+dx\sin\beta} {x^2+y^2} \end{cases} $$ Since $\sin\beta$ is determined (up to the sign) by $\cos\beta$, you have the requested formulas.

Note that the denominators can be rewritten $$ d^2\cos^2\beta+2dh\cos\beta+h^2+d^2\sin^2\beta= d^2+h^2+2dh\cos\beta= x^2+y^2 $$


Mathematica finds that one of the solutions is:

$\alpha = \frac{h x \left(-d^2+h^2+x^2+y^2\right)-\sqrt{-h^2 y^2 \left((d-h)^2-x^2-y^2\right) \left((d+h)^2-x^2-y^2\right)}}{h^2 \left(x^2+y^2\right)}$


$\beta = \tan ^{-1}\left(\frac{-d^2-h^2+x^2+y^2}{2 d h}\right)$.

  • 1
    $\begingroup$ May be a second solution has a plus before radical sign in expression for $\alpha $ and corrosponding negative sign for $\beta $ $\endgroup$ – Narasimham Feb 19 '16 at 22:25



In the second equation, we have:



Thus, we have:


We know that $\cos(\sin^{-1}(\mu))=\sqrt{1-\mu^2}$.








This is as close as I can take you, hopefully someone can build upon this?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.