# Inverse trigonometry prove this

1. Jan 10, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
If $cos^{-1}\frac{x}{a}+cos^{-1}\frac{y}{b} = \alpha$ then show that $\frac{x^2}{a^2}-\frac{2xy}{ab}cos \alpha + \frac{y^2}{b^2} = sin^2 \alpha$

2. Relevant equations

3. The attempt at a solution
I assume inverse functions to be θ and β respectively. So in the second equation for LHS I can convert it like this

$cos^2\theta + cos^2\beta - 2cos \alpha cos\theta cos\beta$

Similarly for RHS
$sin^2 \alpha = sin^2(\theta + \beta)$

2. Jan 12, 2013

### dikmikkel

Taking the inverse cosine of a cosine gives the argument to the inverse cosine. Then you would use the sum of angles trigonometric formula.

3. Jan 12, 2013

### SammyS

Staff Emeritus
So you need to show that $\ \ \cos^2(\theta) + \cos^2(\beta) - 2\cos(\theta + \beta) \cos(\theta)\cos(\beta)= \sin^2(\theta + \beta)\ .$

$\cos(\theta+\beta)=\cos(\theta)\cos( \beta)-\sin(\theta)\sin(\beta)$

$\displaystyle \sin^2(\theta+\beta)=\left(\sin(\theta)\cos(\beta)+\cos(\theta)\sin(\beta) \right)^2\\ \quad\quad\quad=\sin^2(\theta)\cos^2(\beta)+2\sin(\theta)\cos(\beta)\cos(\theta)\sin(\beta)+\cos^2( \theta)\sin^2(\beta)$

So you can get some cancellation.

BTW: Didn't you post this very same question previously?

4. Jan 13, 2013

### utkarshakash

No. Not exactly like this.