proofs

**mortifiedpenguin** · May 27th 2011, 07:14 PM

Cos (x-y)
------------ =
Cos (x+y)

1 + tanx tany
--------------
1 - tanx tany

and

Sin 3x=3sinx-4sin^3x

Any help would be greatly appreciated

**Chris L T521** · May 27th 2011, 07:50 PM

Originally Posted by mortifiedpenguin

Cos (x-y)
------------ =
Cos (x+y)

1 + tanx tany
--------------
1 - tanx tany

Spoiler:

and

Sin 3x=3sinx-4sin^3x

Any help would be greatly appreciated

Spoiler:

I hope these hints help!

**Joanna** · May 27th 2011, 07:56 PM

I would start with

$\frac{1+\tan(x)\tan(y)}{1-\tan(x)\tan(y)}$

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) )

**Soroban** · May 28th 2011, 03:33 AM

Hello, mortifiedpenguin!

$\frac{\cos(x-y)}{\cos(x+y)} \:=\:\frac{1 + \tan x\tan y}{1 - \tan x\tan y}$

We have: . $\frac{\cos(x-y)}{\cos(x+y)} \;=\;\frac{\cos x\cos y + \sin x\sin y}{\cos x\cos y - \sin x\sin y}$

Divide numerator and denominator by $\cos x\cos y$

. . $\dfrac{\dfrac{\cos x\cos y}{\cos x\cos y} + \dfrac{\sin x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y}{\cos x\cos y} - \dfrac{\sin x\sin y}{\cos x\cos y}} \;=\;\frac{1 + \tan x\tan y}{1 - \tan x\tan y}$

**ipower94** · May 28th 2011, 04:16 PM

sin 3x = sin( 2x +x )
= sin2xcosx + sinxcos2x
= (2sinxcosx)cosx + (cos^2x - sin^2x)sinx
= 2sinxcosx^2 + cos^2xsinx - sin^3x
= ...

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