Trigonometry Help (Identities)

Hello, Ruler of Hell!

Quote:

Given: . $\begin{array}{cccc}x\sin^3\!\theta + y\cos^3\!\theta \:= \:\sin\theta\cos\theta & {\color{blue}[1]}\\ x\sin\theta - y\cos\theta \:=\:0 & {\color{blue}[2]}\end{array}$

Prove that: . $x^2 + y^2 \:=\: 1$

From [2], we have: . $y\cos\theta \,=\,x\sin\theta \quad\Rightarrow\quad y \,=\,\frac{x\sin\theta}{\cos\theta}$ .[3]

Substitute into [1]: . $x\sin^3\!\theta + \left(\frac{x\sin\theta}{\cos\theta}\right)\cos^3\ !\theta \:=\:\sin\theta\cos\theta$

. . . . . . . $x\sin^3\!\theta + x\sin\theta\cos^2\!\theta \:=\:\sin\theta\cos\theta$

$\text{Factor: }\;x\sin\theta\underbrace{(\sin^2\!\theta + \cos^2\!\theta)}_{\text{This is 1}} \:=\:\sin\theta\cos\theta$

. . . . . . $x\sin\theta \:=\:\sin\theta\cos\theta \quad\Rightarrow\quad\boxed{ x \:=\:\cos\theta}$

Substitute into [3]: . $y \:=\:\frac{(\cos\theta)(\sin\theta)}{\cos\theta} \quad\Rightarrow\quad\boxed{ y \:=\:\sin\theta}$

Therefore: . $x^2 + y^2 \;=\;\cos^2\!\theta + \sin^2\!\theta \;=\;1$