Solving for $\alpha$ and $\beta$ in the system $x=d\cos(\alpha+\beta)+h\cos\alpha$, $y=d\sin(\alpha+\beta)+h\sin\alpha$

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:
Use:

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

So:

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

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.

• 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. – Ivan Neretin Feb 19 '16 at 22:10
• Near the the fixed point length and angle are $h,\alpha$ ; At mid point $d, (\alpha+\beta)$ – 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:

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

and

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

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

$$\frac{x-h\cos(\alpha)}{d}=\cos(\alpha+\beta)$$

$$\alpha+\beta=\cos^{-1}\left(\frac{x-h\cos(\alpha)}d\right)$$

In the second equation, we have:

$$\frac{y-h\sin(\alpha)}{d}=\sin(\alpha+\beta)$$

$$\alpha+\beta=\sin^{-1}\left(\frac{y-h\sin(\alpha)}d\right)$$

Thus, we have:

$$\cos^{-1}\left(\frac{x-h\cos(\alpha)}d\right)=\sin^{-1}\left(\frac{y-h\sin(\alpha)}d\right)$$

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

$$\frac{x-h\cos(\alpha)}d=\sqrt{1-\left(\frac{y-h\sin(\alpha)}d\right)^2}$$

$$\left(\frac{x-h\cos(\alpha)}d\right)^2=1-\left(\frac{y-h\sin(\alpha)}d\right)^2$$

$$(x-h\cos(\alpha))^2=d^2-(y-h\sin(\alpha))^2$$

$$x^2+y^2-2hx\cos(\alpha)-2hy\sin(\alpha)+h^2\cos^2(\alpha)+h^2\sin^2(\alpha)=d^2$$

$$-2h(x\cos(\alpha)+y\sin(\alpha))+h^2(\cos^2(\alpha)+\sin^2(\alpha))=d^2-x^2-y^2$$

$$-2h(x\cos(\alpha)+y\sin(\alpha))+h^2(1)))=d^2-x^2-y^2$$

$$x\cos(\alpha)+y\sin(\alpha)=\frac{x^2+y^2+h^2-d^2}{2h}$$

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