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Trigonometric Identities - 2147483647 - 2011-01-11
sin(a+b) = sina cosb + cosa sinb
sin(a-b) = sina cosb - cosa sinb
cos(a+b) = cosa cosb - sina sinb
cos(a-b) = cosa cosb + sina sinb
Is there a way to prove these using calculus? (Using derivatives and integrals).
Their form looks a lot like the product rule (xy)' = xy'+x'y. If we take the first one, for example...
int (sina cosb + cosa sinb) = (sina)(cos b)
I'm not sure how to derive the above equation to make it look like sin(a+b).
Trigonometric Identities - Nikkey - 2011-01-11
Why would you prove them using calculus? You can just use basic algebra to do so.
http://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities
Trigonometric Identities - bio9205 - 2011-01-11
Oh, pineapple. 3 months of holidays and my brains are fried. Can't help you on this one. There is a way, but I just can't remember it now.
Trigonometric Identities - 2147483647 - 2011-01-11
Devil's Sunrise Wrote:Why would you prove them using calculus? You can just use basic algebra to do so.
Because it drives me nuts that the form looks so much like the product rule.
Someone help me out please?
Trigonometric Identities - Dusk - 2011-01-11
Your integral is incorrect. You can't integrate wrt a and b at the same time. The trig identities have nothing to do with product rule.