Eliminate the denominator in this expression

[chipotleaway]

Homework Statement

Eliminate the denominator on the RHS of the following expression
sin(\alpha)-sin(\beta) = -\frac{-v(sin(\alpha) cos(\alpha) + sin(\alpha) cos(\beta)}{c-vcos(\alpha)}
and from it, derive
sin(\alpha)-sin(\beta) = -\frac{v}{c}(sin(\alpha)cos(\beta)+sin(\beta)cos(\alpha))

Homework Equations

For a bit of context, this is from a multipart derivation of the relativistic reflection law based on the principle of least time. I'm assuming that you don't need to know what the sines and cosines in terms of the lengths of the triangles because the hint says to 'multiply through' (and substituting back in lengths hasn't made it any better), but I've attached the diagram I'm working from anyway.

The Attempt at a Solution

Some things I've tried:
- worked backwards from the RHS expression in the second equation to see if I could find what to multiply by
- multiplied by top and bottom by c+vcos(\alpha)
- tried writing sin(\alpha) and cos(\alpha) as the ratios of the lengths in hopes of getting a sin(\beta) and [/itex]sin(\alpha)[/itex] back out of it. This is what I got:

-\frac{v}{c}(\frac{x(cos(\alpha)+cos(\beta))}{\sqrt{(d_0+vt_a)^2+x^2)}-v(d_o+vt_a)})

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[Electric Red]

Homework Statement

Eliminate the denominator on the RHS of the following expression
sin(\alpha)-sin(\beta) = -\frac{-v(sin(\alpha) cos(\alpha) + sin(\alpha) cos(\beta)}{c-vcos(\alpha)}
and from it, derive
sin(\alpha)-sin(\beta) = -\frac{v}{c}(sin(\alpha)cos(\beta)+sin(\beta)cos(\alpha))

Homework Equations

For a bit of context, this is from a multipart derivation of the relativistic reflection law based on the principle of least time. I'm assuming that you don't need to know what the sines and cosines in terms of the lengths of the triangles because the hint says to 'multiply through' (and substituting back in lengths hasn't made it any better), but I've attached the diagram I'm working from anyway.

The Attempt at a Solution

Some things I've tried:
- worked backwards from the RHS expression in the second equation to see if I could find what to multiply by
- multiplied by top and bottom by c+vcos(\alpha)
- tried writing sin(\alpha) and cos(\alpha) as the ratios of the lengths in hopes of getting a sin(\beta) and [/itex]sin(\alpha)[/itex] back out of it. This is what I got:

-\frac{v}{c}(\frac{x(cos(\alpha)+cos(\beta))}{\sqrt{(d_0+vt_a)^2+x^2)}-v(d_o+vt_a)})

--

I think I can help you, just tell me, there is missing one bracket on the RHS, could you please tell me where it should be?

 

[ehild]

Homework Statement

Eliminate the denominator on the RHS of the following expression
sin(\alpha)-sin(\beta) = -\frac{-v(sin(\alpha) cos(\alpha) + sin(\alpha) cos(\beta)}{c-vcos(\alpha)}
and from it, derive
sin(\alpha)-sin(\beta) = -\frac{v}{c}(sin(\alpha)cos(\beta)+sin(\beta)cos(\alpha))

There is a parenthesis missing from the end of the RHS of the first equation.

Just multiply the whole equation with the denominator, expand and simplify.


ehild

 

[chipotleaway]

Ah I got it now, thanks ehild and Electric Red.

Somewhere along the line, I got the idea that to solve an equation we weren't allowed to multiply it by anything other than one or else we change it somehow...though it makes sense now because if we multiply both sides by the same factor then it'll just cancel in the end...

 

[ehild]

Ah I got it now, thanks ehild and Electric Red.

Somewhere along the line, I got the idea that to solve an equation we weren't allowed to multiply it by anything other than one or else we change it somehow...though it makes sense now because if we multiply both sides by the same factor then it'll just cancel in the end...

When you multiply with an expression you have to exclude the cases when it is zero. So you say: Multiply by c-vcos(α) ≠ 0. Here it is sure as v∠c and cos(α) ≤1.

ehild