Eliminating parameter $\beta$ from $x=\cos 3 \beta + \sin 3 \beta$, $y = \cos \beta - \sin \beta$

Question

Based on the given parametric equations:

$$\begin{align} x &=\cos 3 \beta + \sin 3 \beta \\ y &= \cos \beta \phantom{3}- \sin \beta \end{align}$$

Eliminate the parameter $\beta$ to prove that $x-3y+2y^3=0$.

What I got so far:

$$\cos 3 \beta + \sin 3 \beta = ( \cos \beta - \sin \beta)(1+4\sin\beta\cos\beta)$$

Which trigonometric identity should I use to proceed?

Answer 1

Hint:

$$\cos3\beta+\sin3\beta=4(\cos^3\beta-\sin^3\beta)-3(\cos\beta-\sin\beta)$$

$$(\cos\beta-\sin\beta)^3=\cos^3\beta-\sin^3\beta-3\cos\beta\sin\beta(\cos\beta-\sin\beta)$$

$$y^2=?$$

Replace the values of $\cos\beta\sin\beta,\cos\beta-\sin\beta$

Answer 2

Hint: Write your system in the form $$x=\sqrt{2}\sin\left(\frac{\pi}{4}-\beta\right)(1+2\sin(2\beta))$$ $$y=\sqrt{2}\sin\left(\frac{\pi}{4}-\beta\right)$$ then you will get $$x=y(1+2\sin(2\beta))$$ and you can eliminate $\beta$

Answer 3

$$y=\sqrt2\cos\left(\dfrac\pi4+\beta\right)=\sqrt2\cos t$$ where $\dfrac\pi4+\beta=t$

$$x= \cos3\left(t-\dfrac\pi4\right)+\sin3\left(t-\dfrac\pi4\right)$$

As $\cos3\dfrac{\pi}4=\cos\left(\pi-\dfrac\pi4\right)=-\cos\dfrac\pi4=?$

$\sin3\dfrac\pi4=?$

$$x=-\dfrac1{\sqrt2}\cos3t+\dfrac1{\sqrt2}\cdot\sin3t-\dfrac1{\sqrt2}\cdot\sin3t-\dfrac1{\sqrt2}\cdot\cos3t$$

$$\implies\sqrt2x=-2\cos3t$$

Use $\cos3t=4\cos^3t-3\cos t$

Answer 4

Making

$$ y = \frac 12\left(e^{i\beta}+e^{-i \beta}\right)-\frac{1}{2i}\left(e^{i\beta}-e^{-i\beta}\right) $$

we have after powering and collecting

$$ y^3 = \frac 32\left(\cos\beta-\sin\beta\right)-\frac 12\left(\cos(3\beta)+\sin(3\beta)\right) $$

then follows

$$ y^3 = \frac 32 y - \frac 12 x $$