Eliminating $\theta$ from $a\cos(\theta-\alpha)=x$ and $b\cos(\theta-\beta)=y$

Question

Eliminate $\theta$ in following equations $$\begin{align} a \cos(\theta-\alpha) &= x \\ b \cos(\theta- \beta) &=y \end{align}$$

I am trying to solve this problem but still I am unable to get the perfect answer I added both the equations but it transformed it to $2 \cos(\theta+(\alpha + \beta)/2)$

Answer 1

\begin{align} \frac{x}{a}+\frac{y}{b} &= \cos (\theta-\alpha)+\cos (\theta-\beta) \\ &=2\cos \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta-2\theta}{2} \\ \frac{x}{a}-\frac{y}{b} &= \cos (\theta-\alpha)-\cos (\theta-\beta) \\ &=-2\sin \frac{\alpha-\beta}{2} \sin \frac{\alpha+\beta-2\theta}{2} \\ 1 &= \left( \frac{\frac{x}{a}+\frac{y}{b}}{2\cos \frac{\alpha-\beta}{2}} \right)^2+ \left( -\frac{\frac{x}{a}-\frac{y}{b}}{2\sin \frac{\alpha-\beta}{2}} \right)^2 \\ \sin^2 (\alpha-\beta) &= \frac{x^2}{a^2}-\frac{2xy\cos (\alpha-\beta)}{ab}+\frac{y^2}{b^2} \end{align}

• The curve is known as Lissajous figure (with same frequencies).

• The area bounded by the ellipse is $A=\pi ab\sin (\beta-\alpha)$.

• $A>0$ gives anti-clockwise trace whereas $A<0$ for clockwise.

• When the curve degenerates to a line segment, $A=0$

Answer 2

Hint:

$$\cos(u+v)=\cos(u)\cos(v)-\sin(u)\sin(v)$$

Answer 3

For simplicity, define $u:=x/a$ and $v:=y/b$, so that we have $$\begin{align} u &= \cos(\theta-\alpha) = \cos\theta \cos\alpha + \sin\theta\sin\alpha \\ v &= \cos(\theta-\beta) = \cos\theta\cos\beta + \sin\theta\sin\beta \end{align}$$ This is a linear system in $\cos\theta$ and $\sin\theta$. Solving, we obtain $$\cos\theta = \frac{v\sin\alpha - u \sin\beta}{ \sin\alpha \cos\beta - \cos\alpha\sin\beta} = \frac{v\sin\alpha-u\sin\beta}{\sin(\alpha-\beta)} \qquad\qquad \sin\theta = \frac{u \cos\beta - v \cos\alpha}{\sin(\alpha-\beta)}$$ Then, because $\cos^2\theta + \sin^2\theta = 1$, we can write $$\frac{\left(v\sin\alpha-u\sin\beta\right)^2}{\sin^2(\alpha-\beta)} + \frac{\left(u\cos\beta-v\cos\alpha\right)^2}{\sin^2(\alpha-\beta)} = 1$$ so that $$u^2 + v^2 - 2 u v (\sin\alpha\sin\beta+\cos\alpha\cos\beta) = \sin^2(\alpha-\beta)$$ which becomes

$$u^2 + v^2 - 2 u v \cos(\alpha-\beta) = \sin^2(\alpha-\beta)$$