- #1

chipotleaway

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## Homework Statement

Show that

[tex]sin(\alpha)-sin(\beta)=-\frac{v}{c}sin(\alpha+\beta)[/tex]

is equivalent to

[tex]cos(\beta)=\frac{-2\frac{v}{c}+(1+\frac{v^2}{c^2})cos(\alpha)}{1-2\frac{v}{c}cos(\alpha)+\frac{v^2}{c^2}}[/tex]

## Homework Equations

This is the last step in the derivation of the relativistic reflection law. The first equation was derived using the principle of least time and the second is from Einstein's paper on special relativity.

[itex]\alpha[/itex] = angle of incidence

[itex]\beta[/itex] = angle of reflection

[itex]v[/itex] = velocity of mirror away from light

[itex]c[/itex] = speed of light

link to diagram

## The Attempt at a Solution

I used the identity [itex]sin(\alpha+\beta = sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)[/itex] on the RHS, squared both sides, changed by the squared sines into cosines squared - some more algebra and this is the best I've got at the moment:

[tex]2c^2(1-sin(\alpha)sin(\beta))-(c^2-v^2)(cos^2(\alpha)+cos^2(\beta))=-2v^2cos^2(\alpha)cos^2(\beta)+2sin(\alpha)cos(\beta)sin(\beta)cos( \alpha)[/tex]

How do I get rid of the sines?