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You need to make use of the formulae for sin(A+B) and cos(A+B), and that cos is an even function:

sin(A+B) = cos A sin B + sin A cos B

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

cos even fn → cos(-x) = cos(x)

To prove: (cos A + sin A)(cos 2A + sin 2A) = cos A + sin 3A

The steps are to work with the left hand side, expand the brackets, collect [useful] terms together, apply A+B formula above (backwards) and apply even nature of cos function:

(cos A + sin A)(cos 2A + sin 2A) = cos A cos 2A + cos A sin 2A + sin A cos 2A + sin A sin 2A

= (cos A cos 2A + sin A sin 2A) + (cos A sin 2A + sin A cos 2A)

= cos(A - 2A) + sin(A + 2A)

= cos(-A) + sin 3A

= cos A + sin 3A

which is the right hand side as required.

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โˆ™ 2015-09-09 20:54:30
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โˆ™ 2015-09-09 15:26:16

As usual, some signs got lost in the question. I would start by applying the double-angle and triple-angle formulae, to express everything in terms of sin(A) and cos(A).

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Q: How would you prove left cosA plus sinA right times left cos2A plus sin2A right equals cosA plus sin3A?
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