Proving Trigonometric Identity: Double Angle Formula

[likearollings]

I am not quite sure how to prove this,

I can see it involves using a double angle formula (see http://mathworld.wolfram.com/Double-AngleFormulas.html).

I have tried working backwards but to no avail.

Any ideas?

http://img847.imageshack.us/img847/9428/trig.png

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[AlephZero]

Use
\sin a \cos b = (\sin (a+b) + \sin(a-b))/2[/itex]

 

[likearollings]

Thanks! I will give that a go

 

[likearollings]

Use
\sin a \cos b = (\sin (a+b) + \sin(a-b))/2[/itex]

<br /> <br /> Thanks, I managed to do it using <br /> <br /> &lt;br /&gt; \cos a \sin b = (\sin (a+b) - \sin(a-b))/2&lt;br /&gt;<br /> <br /> PROBLEM SOLVED!