# Math Help - A Couple Trig Identities

1. ## A Couple Trig Identities

This is the beginning of a couple of identities that i will probably need assistance on, all help given is greatly appreciated.

$\displaystyle{\frac{\tan x - \cot x}{\tan x + \cot x}=2\sin^2x-1}$
first i switched them all to sin and cos
then i found like denominators and combined
then i multiplied by the reciprocal leading me to here:

$\displaystyle{\frac{\sin x - \cos x}{\cos x \sin x} \bullet \frac{\cos x \sin x}{\sin x + \cos x}}$

then got $\displaystyle{\frac{\sin x - \cos x}{\sin x + \cos x} = 2\sin^2x-1}$
I'm unsure where to go from here

2. Originally Posted by ~berserk
This is the beginning of a couple of identities that i will probably need assistance on, all help given is greatly appreciated.

$\displaystyle{\frac{\tan x - \cot x}{\tan x + \cot x}=2\sin^2x-1}$
first i switched them all to sin and cos
then i found like denominators and combined
then i multiplied by the reciprocal leading me to here:

$\displaystyle{\frac{\sin x - \cos x}{\cos x \sin x} \bullet \frac{\cos x \sin x}{\sin x + \cos x}}$

then got $\displaystyle{\frac{\sin x - \cos x}{\sin x + \cos x} = 2\sin^2x-1}$
I'm unsure where to go from here
If you did the 1st one that way, your squares have gone missing...

$\displaystyle\frac{\left[\frac{sinx}{cosx}-\frac{cosx}{sinx}\right]}{\left[\frac{sinx}{cosx}+\frac{cosx}{sinx}\right]}=\frac{\left[\frac{1}{sinxcosx}\right]}{\left[\frac{1}{sinxcosx}\right]}\;\frac{sin^2x-cos^2x}{sin^2x+cos^2x}$

$=sin^2x-cos^2x$

since $sin^2x+cos^2x=1$

and now use $cos^2x=1-sin^2x$ to finish

3. also there is this one where I am unsure where to start: $(\sin x + \cos x)^2 = 1 +\sin 2x$

4. Originally Posted by ~berserk
also there is this one where I am unsure where to start: $(\sin x + \cos x)^2 = 1 +\sin 2x$
$(\sin x + \cos x)^2 = \left(\sin^2{x}+\cos^2{x}\right)+2\sin{x}\cos{x} = ...$

5. Originally Posted by ~berserk
also there is this one where I am unsure where to start: $(\sin x + \cos x)^2 = 1 +\sin 2x$
That's another $sin^2x+cos^2x=1$ identity in disguise.

That's a crucial one to remember!

$(sinx+cosx)(sinx+cosx)=sinx(sinx+cosx)+cosx(sinx+c osx)$

$=sin^2x+sinxcosx+cosxsinx+cos^2x=sin^2x+cos^2x+2si nxcosx$

Finish up with the identity $sin2x=2sinxcosx$