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

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

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

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Behaviour, Hakim, Daniel Fischer, userNaN, Jack D'Aurizio
If this question can be reworded to fit the rules in the help center, please edit the 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

5 Answers 5

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

share|improve this answer

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

share|improve this answer

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

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

cos(A+B)= cosAcosB - sinAsinB and cos(A-B)= cosAcosB + sinAsinB

therefore cos(A−B)−cos(A+B)=cosAcosB + sinAsinB - (cosAcosB - sinAsinB)

=cosAcosB + sinAsinB - cosAcosB + sinAsinB = 2sinAsinB.

hence proved.

also tidet please accept an answer.

share|improve this answer

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