Prove the composite argument property for cos(A-B)

**wiseguy** · July 22nd 2010, 02:08 PM

"Show numerically that the composite argument property for cos (A - B) is correct by substituting 37° for A and 21° for B and showing that you get the same answer for both sides of the equation."

My response is:

$cos(37°-21°)=cos(16°)=-0.9576594803<br /> <br /> cos(37°-21°)=cos(37°)-cos(21°) does not equal -0.9576594803$

I'm just wondering if I answered the question... the point is that cosine does not distribute.

**pickslides** · July 22nd 2010, 02:26 PM

I think you are supposed to use the fact that $\cos(a-b) = \cos a \cos b +\sin a \sin b$

**wiseguy** · July 22nd 2010, 02:31 PM

So $cos(37-21)=cos37cos21+sin37sin21$ would suffice?

**pickslides** · July 22nd 2010, 02:40 PM

That's my understanding of the question.

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