Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
1  
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
add comment

4 Answers

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

share|improve this answer
add comment

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.