# Thread: Help With Trig Equation

1. ## Help With Trig Equation

if: xsin(A) + ycos(A) = R and xcos(A) - ysin(A)= Q

Find the value of (x^2) + (y^2)

thanks

2. Hello, white_cap!

If . $\begin{array}{ccc}x\sin(A) + y\cos(A) &= &R \\ x\cos(A) - y\sin(A)& =& Q\end{array}$

find the value of: . $x^2 + y^2$

Square both equations:

$[x\sin(A)\;+\;y\cos(A)]^2 \;=\;R^2 \quad \Rightarrow$ . . $x^2\sin^2(A)\;+\;2xy\sin(A)\cos(A)\;+\;y^2\cos^2(A ) \;=\;R^2$

$[x\cos(A)\;-\;y\sin(A)]^2\;=\;Q^2 \quad\Rightarrow$ . . $x^2\cos^2(A)\;-\;2xy\sin(A)\cos(A)\;+\;y^2\sin^2(A) \;=\;Q^2$

Add the equations: . $x^2\!\cdot\!\underbrace{\left[\sin^2(A) + \cos^2(A)\right]}_{\text{This is 1}} +\: y^2\!\cdot\!\underbrace{\left[\sin^2(A) + \cos^2(A)\right])}_{\text{This is 1}} \;=\;R^2+Q^2$

Therefore: . $x^2 + y^2 \;=\;R^2+Q^2$

3. Wow that was brilliant. I understand it, but how did (would) you ever think of adding both equations, unless you wrote them above each other and realised that it would 'work out' nicely?

4. Hello again, white_cap!

It's one of those thing we learn to watch for.

When I see both $\sin\theta$ and $\cos\theta$, that technique come to mind

It turns up regularly in Parametric Equations.

Example: . $\begin{array}{ccc}x & = & 3\sin\theta \\ y & = & 2\cos\theta\end{array}$

Eliminate the parameter $\theta$.
Get an equation involving $x$ and $y$ (only).

We have: . $\begin{Bmatrix}\dfrac{x}{3} \:=\:\sin\theta \\ \\ \dfrac{y}{2} \:=\:\cos\theta \end{Bmatrix}$
Square the equations: . $\begin{Bmatrix}\dfrac{x^2}{9} \:=\:\sin^2\theta \\ \\ \dfrac{y^2}{4} \:=\:\cos^2\theta \end{Bmatrix}$

Add the equations: . $\frac{x^2}{9} + \frac{y^2}{4} \:=\:\underbrace{\sin^2\theta + \cos^2\theta}_{\text{This is 1}}$

Therefore: . $\frac{x^2}{9} + \frac{y^2}{4} \:=\:1$ . . . . There!