Trigonometry Help (Identities)

**Ruler of Hell** · January 6th 2009, 05:11 AM

Hi!
I just have one sum I need.
xsin³θ + ycos³θ = sinθ.cosθ AND xsinθ - ycosθ = 0

Prove that: x² + y² = 1

I tried a lot, but couldn't get it.

Thanks in advance.

**Soroban** · January 6th 2009, 05:46 AM

Hello, Ruler of Hell!

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

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