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_____ cos b = 1/2[sin(a+b) + sin(a-b)] A. sin a B. sin b C. cos a D. cos b ***my answer; A. sin a is that right? @terenzreignz :)
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hmm... I hate these identities. I don't recall when I ever had to use them...
Let's see... sin(a+b) = sin(a)cos(b) + cos(a)sin(b) sin(a-b) = sin(a)cos(b) - cos(a)sin(b) thus sin(a+b) + sin(a-b) = 2sin(a)cos(b) halving that, you get sin(a)cos(b) how about that, eh? :D
ooh!! yay!! :D thanks!! :D haha woohoo! :D
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