# Tough Trig Identity

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• Jan 27th 2014, 06:22 PM
sakonpure6
Tough Trig Identity
cot(2x) = csc(2x) - tanx

I tried working with either side, but I couldn't get anywhere. Any help is appreciated!!
• Jan 27th 2014, 06:36 PM
romsek
Re: Tough Trig Identity
Quote:

Originally Posted by sakonpure6
cot(2x) = csc(2x) - tanx

I tried working with either side, but I couldn't get anywhere. Any help is appreciated!!

couldn't get anywhere? why?

$\cot(2x)=\frac{\cos(2x)}{\sin(2x)}=\frac{2\cos^2(x )-1}{2\sin(x) \cos(x)}$

$\csc(2x)=\frac{1}{\sin(2x)}=\frac{1}{2\sin(x) \cos(x)}$

you can finish it from here
• Jan 27th 2014, 06:45 PM
SlipEternal
Re: Tough Trig Identity
Start with the LHS.
Use the following identities:

\begin{align*}\cot x & = \dfrac{\cos x}{\sin x} \\ \cos (2x) & = 1-2\sin^2 x \\ \dfrac{a-b}{c} & = \dfrac{a}{c} - \dfrac{b}{c}\\ \dfrac{1}{\sin x} & = \csc x \\ \sin(2x) & = 2\sin x \cos x \\ \dfrac{\sin x}{\cos x} & = \tan x\end{align*}
• Jan 27th 2014, 06:47 PM
SlipEternal
Re: Tough Trig Identity
Quote:

Originally Posted by romsek
couldn't get anywhere? why?

$\cot(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{2\sin(x) \cos(x)}{2\cos^2(x)-1}$

Should be: $\cot(2x) = \dfrac{\cos(2x)}{\sin (2x)} = \dfrac{2\cos^2 (x)-1}{2\sin (x) \cos (x)}$
• Jan 27th 2014, 08:25 PM
ibdutt
Re: Tough Trig Identity