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Trigonometry
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(cos^3x - sin^3x)/(cosx - sinx) = 1 + sinx cosx; Verify the Identity. I am not sure how to get to the answer. Any help getting me started would be greatly appreciated.
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You should recognize the form of the numerator. A difference of cubes. This can be factored.
Can you factor out a sincos to then create sincos(sin^2 +cos^2
\(a^3-b^3=(a-b)(a^2+ab+b^2)\)
As tkhunny mentioned, you have something that looks like \[\frac{a^3-b^3}{a-b}\] Using the difference of cubes pattern, you can factor to get \[\frac{(a-b)(a^2+ab+b^2)}{(a-b)}\]
Thank you very much... let me see what I do next, but that was very cool...
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Thank you for the help on this one... I got it...
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