if cos x+ cos y = a , sin x +sin y = b show cos(x-y) in terms of a and b..
cos(s-t) = cos s cos t + sin s sin t Does that help?
i don't know what to do after that..
cos(x-y) = cos x (a - cos x) + sin x (b - sin x)
but the answer is \[(a ^{2}+b^{2}-2)\div2\]
so = a cos x - cos^2 x + b sin x - sin^2 x = a cos x + b sin x - 1 Sorry, someone else will have show me how I got off track with this one.
its okay.. thank you anyway..:)
Try working backword from the answer.
is cos x + cos y is the same as cos(x+y)?
No. But I got it now. Let me do it backward..
1/2(a^2 + b^2 - 2) where a^2 = cos^2 x + 2 cos x cos y + cos^2 y b^2 = sin^2 x + 2 sin x siny + sin^2 y so a^2 + b^2 - 2 = 2 cos x cos y + 2 sin x siny Multiply by 1/2 we have cos x cos y + sin x sin y which equals cos (x-y)
ouhhh.. okie..
Yeah. I think it's right but I'd never have got there in the forward direction.
im trying to do in forward direction based from the backward direction you showed..
i think i get it..
Great!
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