Identity manipulation but a tough one!

**dojo** · November 2nd 2010, 03:26 PM

Hi, I'm struggling with this question:

Show that

$4sin(\alpha + \beta)sin(\alpha + \gamma)sin(\beta + \gamma) + sin(2\alpha + 2\beta + 2\gamma) = sin(2\alpha) + sin(2\beta) + sin(2\gamma)$

I have covered identities but I get in a huge expansion that is so confusing and from past questions tend to have a neater way of doing things.

Your help would be greatly appreciated.
D

**Opalg** · November 3rd 2010, 02:43 AM

Originally Posted by dojo

Show that

$4sin(\alpha + \beta)sin(\alpha + \gamma)sin(\beta + \gamma) + sin(2\alpha + 2\beta + 2\gamma) = sin(2\alpha) + sin(2\beta) + sin(2\gamma)$

I have covered identities but I get in a huge expansion that is so confusing and from past questions tend to have a neater way of doing things.

I think that the tidiest way to do this is to use the addition formulas

$(1)\;\sin x + \sin y = 2\sin\frac12(x+y)\cos\frac12(x-y),$
$(2)\;\sin x - \sin y = 2\cos\frac12(x+y)\sin\frac12(x-y),$
$(3)\;\cos x - \cos y = 2\sin\frac12(x+y)\sin\frac12(x-y).$

Use those formulas to get

$(4)\; \sin2(\alpha+\beta+\gamma) - \sin2\alpha = 2\cos(2\alpha+\beta+\gamma)\sin(\beta+\gamma)$ (from (2)),
$(5)\; \sin2\beta + \sin2\gamma = 2\sin(\beta+\gamma)\cos(\beta - \gamma)$ (from (1)),

and therefore

$(6)\; \sin2(\alpha+\beta+\gamma) - \sin2\alpha - \sin2\beta - \sin2\gamma = 2\sin(\beta+\gamma)\bigl(\cos(2\alpha+\beta+\gamma ) - \cos(\beta - \gamma)\bigr)$ (subtracting (5) from (4)).

Also,

$(7)\; \cos(2\alpha+\beta+\gamma) - \cos(\beta - \gamma) = 2\sin(\alpha+\beta)\sin(\alpha+\gamma)$ (from (3)).

Finally, substitute the result from (7) into (6) to get your formula.

The interesting thing here is that although the formula is completely symmetrical in $\alpha$ , $\beta$ and $\gamma$ , it is necessary to break the symmetry in order to get an efficient proof (in this case, by treating the term $\sin2\alpha$ differently from its "partners" $\sin2\beta$ and $\sin2\gamma$ ).

**dojo** · November 5th 2010, 09:10 AM

Thanks for running this through - its a great help and I feel a lot more confident with this type of question.

However, when I rearrange and replace as you suggest I get:

$4sin(\alpha + \beta)sin(\alpha + \gamma)sin(\beta + \gamma) = - 4sin(\alpha + \beta)sin(\alpha + \gamma)sin(\beta + \gamma)$ which isn't true.

Where am I going wrong??

**Unknown008** · November 5th 2010, 11:05 AM

$\cos x - \cos y = 2\sin\frac12(x+y)\sin\frac12(x-y)$

This line should be:

$\cos x - \cos y = -2\sin\frac12(x+y)\sin\frac12(x-y)$

This makes:

$\cos(2\alpha+\beta+\gamma) - \cos(\beta - \gamma) = -2\sin(\alpha+\beta)\sin(\alpha+\gamma)$

Otherwise, it's a very interesting problem

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