Cos (x-y)
------------ =
Cos (x+y)
1 + tanx tany
--------------
1 - tanx tany
and
Sin 3x=3sinx-4sin^3x
Any help would be greatly appreciated
I would start with
$\displaystyle \frac{1+\tan(x)\tan(y)}{1-\tan(x)\tan(y)}$
and transform them into something to which you can apply the angle sum identities for cosine. (i.e. cos(x+y) = cos(x)cos(y)-sin(x)sin(y) and
cos(x-y)=cos(x)cos(y)+sin(x)sin(y) )
Hello, mortifiedpenguin!
$\displaystyle \frac{\cos(x-y)}{\cos(x+y)} \:=\:\frac{1 + \tan x\tan y}{1 - \tan x\tan y}$
We have: .$\displaystyle \frac{\cos(x-y)}{\cos(x+y)} \;=\;\frac{\cos x\cos y + \sin x\sin y}{\cos x\cos y - \sin x\sin y}$
Divide numerator and denominator by $\displaystyle \cos x\cos y$
. . $\displaystyle \dfrac{\dfrac{\cos x\cos y}{\cos x\cos y} + \dfrac{\sin x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y}{\cos x\cos y} - \dfrac{\sin x\sin y}{\cos x\cos y}} \;=\;\frac{1 + \tan x\tan y}{1 - \tan x\tan y} $
sin 3x = sin( 2x +x )
= sin2xcosx + sinxcos2x
= (2sinxcosx)cosx + (cos^2x - sin^2x)sinx
= 2sinxcosx^2 + cos^2xsinx - sin^3x
= ...