systems of equationstrigonometry
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?
Note that $$x^2 + y^2 = \sin^2(\theta/2) + \cos^2(\theta/2) = 1$$
\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, $A=0$
Best Answer
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 $\cos\beta\sin\beta,\cos\beta-\sin\beta$