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Trigonometry
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Simplify cos(x-y)+cos(x+y)/cosx
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is the whole thing over cos(x) or is it just the cos(x+y) that is over cos(x)?
The whole thing is over cosx
okay do you know your sum and difference formulae for cos(x+y) and cos(x-y)?
Yes
so that's the first step. expand the top 2 terms like so: \[\frac{\cos(x-y)+\cos(x-y)}{\cos(x)}=\] \[\frac{(\cos(x)\cos(y)+\sin(x)\sin(y))+(\cos(x)\cos(y)-\sin(x)\sin(y))}{\cos(x)}=\] the sin terms cancel out and the cos terms add together to give you: \[\frac{2\cos(x)\cos(y)}{\cos(x)}=2\cos(y)\] and that is your answer :)
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oops the first part is actually: \[\frac{\cos(x-y)+\cos(x+y)}{\cos(x)}\] sorry :)
Thanks very much
use tis identity 2cos x cos y = cos (x + y) + cos (x – y) 2cosxcosy/cosx=2cosy
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