# proofs - help

• May 27th 2011, 07:14 PM
mortifiedpenguin
proofs - help
Cos (x-y)
------------ =
Cos (x+y)

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

and

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

Any help would be greatly appreciated (Nod)
• May 27th 2011, 07:50 PM
Chris L T521
Quote:

Originally Posted by mortifiedpenguin
Cos (x-y)
------------ =
Cos (x+y)

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

Spoiler:
$\displaystyle \frac{1+\tan x\tan y}{1-\tan x\tan y} = \frac{1+\tan x\tan y}{1-\tan x\tan y}\cdot\frac{\cos x\cos y}{\cos x\cos y}=\ldots$

Quote:

and

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

Any help would be greatly appreciated (Nod)
Spoiler:
\displaystyle \begin{aligned}\sin(3x)&=\sin(2x)\cos x +\cos(2x)\sin x\\&=2\sin x\cos^2x+(1-2\sin^2x)\sin x\\&=\ldots\end{aligned}

Continue applying identities to get everything in terms of sine and its powers. Then simplify to get the result!

I hope these hints help!
• May 27th 2011, 07:56 PM
Joanna

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

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) )
• May 28th 2011, 03:33 AM
Soroban
Hello, mortifiedpenguin!

Quote:

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

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 $\displaystyle \cos x\cos y$

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

• May 28th 2011, 04:16 PM
ipower94
sin 3x=3sinx - 4sin^3x
sin 3x = sin( 2x +x )
= sin2xcosx + sinxcos2x
= (2sinxcosx)cosx + (cos^2x - sin^2x)sinx
= 2sinxcosx^2 + cos^2xsinx - sin^3x
= ...