# Math Help - Trig Identity

1. ## Trig Identity

(1+cosx)/sinx = cot(x/2)
Very tough... I am thinking now of working with the right side and doing :

cot(x/2) * cot(2x)/cot(2x)
= cot^2x/cot2x // Is this true?

because I have no idea what to do when I am given some thing like (x/2)

Any hints are really appreciated

2. ## Re: Trig Identity

Hello, sakonpure6!

You are expected to know these double-angle identities:

. . $\cos^2\theta \:=\:\frac{1+\cos2\theta}{2} \quad\Rightarrow\quad 1 + \cos x \:=\:2\cos^2\tfrac{x}{2}$

. . $\sin2\theta \:=\:2\sin\theta\cos\theta \quad\Rightarrow\quad \sin x \:=\:\2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$

$\text{Prove: }\:\frac{1+\cos x}{\sin x} \:=\:\cot\tfrac{x}{2}$

We have: . $\frac{1+\cos x}{\sin x} \;=\;\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2} \cos\frac{x}{2}} \;=\;\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} \;=\;\cot\tfrac{x}{2}$

3. ## Re: Trig Identity

Do you know the "sum" formulas: sin(A+ B)= sin(A)cos(B)+ cos(A)sin(B) and cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)?
Setting A= B= X gives the "double angle" formulas: sin(2X)= 2sin(X)cos(X) and cos(2X)= cos^2(X)- sin^2(X)

Letting Y= 2X in that last lets us derive the "half angle" formulas:
$cos(Y)= cos^2(Y/2)- sin^2(Y/2)= cos^2(Y/2)- (1- cos^2(Y/2))= 2cos^2(Y/2)- 1$
$2cos^2(Y/2)= cos(Y)+ 1$
$cos(Y/2)= \dfrac{\sqrt{(cos(Y)+ 1)}}{2}$

$cos(Y)= cos^(Y/2)- sin^2(Y/2)= 1- sin^2(Y/2)- sin^2(Y/2)= 1- 2sin^2(Y/2)$
$2sin^2(Y/2)= 1- cos(Y)$
$sin(Y/2)= \sqrt{(1- cos(Y))/2}$

And, of course, $tan(x/2)= \frac{sin(x/2)}{cos(x/2)}$ and $cos(x/2)= \frac{cos(x/2)}{sin(x/2)}$