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Show that $\cos(A-B)-\cos(A+B)=2\sin A\sin B.$

Using compound angle and identity for $\cos.$
Anybody have an idea how to solve this question?

Cheers all...

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closed as off-topic by ᴡᴏʀᴅs, Hakim, Daniel Fischer, Norbert, Jack D'Aurizio Jul 5 '14 at 19:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

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What do you mean by "compound angle and identity for cosine"? Do you know an identity for $\cos(A+B)$? – Gerry Myerson Mar 19 '13 at 23:53
@Peter, "indentity"? – Gerry Myerson Mar 20 '13 at 0:08
@Gerry, when editing questions I try not to change the OP's wording, only the Latex. It's a question of should I impose my own vocabulary on the OP. – Peter Phipps Mar 20 '13 at 0:12
@Peter, you changed --- indeed, improved --- OP's spelling. You replaced some horrible thing with "compound", and you replaced "indenty" with $indentity". You made several corrections, but you missed one. – Gerry Myerson Mar 20 '13 at 0:19
@Gerry, I spelt identity wrongly. My apologies. – Peter Phipps Mar 20 '13 at 0:30

$$\cos(A-B)-\cos(A+B)=\cos A\cos B+\sin A\sin B-(\cos A\cos B-\sin A\sin B)$$

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I suspect that you're being asked to use the angle sum/difference formulas for cosine.

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cos(A−B)−cos(A+B)=cos(A)cos(B)+sin(A)sin(B)−(cos(A)cos(B)−sin(A)sin(B))= 2sin(A)cos(B) by the identity for the cosine of a sum and difference of its arguments.

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Identities for the cosine of sum and difference of two angles: $$ \begin{align} \cos(A-B)&=\cos(A)\cos(B)+\sin(A)\sin(B)\\ \cos(A+B)&=\cos(A)\cos(B)-\sin(A)\sin(B) \end{align} $$ Subtract.

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