Inverse trigonometry prove this

[utkarshakash]

Homework Statement

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

Homework Equations

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)

 

[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.

 

[SammyS]

Homework Statement

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

Homework Equations

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)

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\\<br /> \quad\quad\quad=\sin^2(\theta)\cos^2(\beta)+2\sin(\theta)\cos(\beta)\cos(\theta)\sin(\beta)+\cos^2( \theta)\sin^2(\beta)<br />

So you can get some cancellation.


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

 

[utkarshakash]

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\\<br /> \quad\quad\quad=\sin^2(\theta)\cos^2(\beta)+2\sin(\theta)\cos(\beta)\cos(\theta)\sin(\beta)+\cos^2( \theta)\sin^2(\beta)<br />

So you can get some cancellation.


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

No. Not exactly like this.