Asked by Alex
Okay, I've been getting some of these, but I can't seem to verify this identity... any help? Here's the problem
Sin(x+y) + Sin(x-y) = 2sinxcosy
Okay, I've been working on the left side, and distribute, getting:
Sinx + Siny + Sinx - Siny
And, the sinx's add up to the 2sinx that I need for the right side, but the siny's cancel out if I don't change them around. So I changed one of them to 1/cscy... but I can't seem to work with that and the other siny to end up with cosy.
Where did I go wrong, or where do I go with it now?
Sin(x+y) + Sin(x-y) = 2sinxcosy
Okay, I've been working on the left side, and distribute, getting:
Sinx + Siny + Sinx - Siny
And, the sinx's add up to the 2sinx that I need for the right side, but the siny's cancel out if I don't change them around. So I changed one of them to 1/cscy... but I can't seem to work with that and the other siny to end up with cosy.
Where did I go wrong, or where do I go with it now?
Answers
Answered by
Count Iblis
Sin(x+y) = Sin(x)Cos(y) + Cos(x)Sin(y)
If you replace y by minus y and add you get twice the first term, because
cos(-y) = cos(y) and sin(-y) = - sin(y).
If you replace y by minus y and add you get twice the first term, because
cos(-y) = cos(y) and sin(-y) = - sin(y).
Answered by
Alex
oh, dang, forgot about that identity!
Okay, well, adding that up it ends up as 2sinx2cosy... but I need it to be 2sinx(1)cosy.
or am I really tired and I'm thinking too out of the box to realize the entire term of 2sinxcosy consists of 2sinx's and 2cosy's?
if not, what do I switch up to drop a cosy to satisfy the right side of the equation? oh man, am i sure forgetting everything about math, lol.
Okay, well, adding that up it ends up as 2sinx2cosy... but I need it to be 2sinx(1)cosy.
or am I really tired and I'm thinking too out of the box to realize the entire term of 2sinxcosy consists of 2sinx's and 2cosy's?
if not, what do I switch up to drop a cosy to satisfy the right side of the equation? oh man, am i sure forgetting everything about math, lol.
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