# Thread: Simplifying trig expression

1. ## Simplifying trig expression

I have to simplify (or get it in terms of tan I guess?) $\cot (\frac{2\pi }{3} - x)$

I'm not sure how to get the reference angle and subtract the angle 'x' from it to get an expressional value...how would I do this?

2. ## Re: Simplifying trig expression

Originally Posted by daigo
I have to simplify (or get it in terms of tan I guess?) $\cot (\frac{2\pi }{3} - x)$

I'm not sure how to get the reference angle and subtract the angle 'x' from it to get an expressional value...how would I do this?
tan2pi/3==-root3 I'm writing t for tanx

tan(2pi/3-x) = (-root3-t)/(1+root3*t) So cot (2pi/3-x) = (1+root3*t)/(-root3-t) = -(1+root3*t)/root3+t)

3. ## Re: Simplifying trig expression

note the cofunction identity ...

$\cot{t} = \tan\left(\frac{\pi}{2} - t\right)$

so ...

$\cot\left(\frac{2\pi}{3} - x\right) = \tan\left[\frac{\pi}{2} - \left(\frac{2\pi}{3} - x\right)\right] = \tan\left(x -\frac{\pi}{6}\right)$

4. ## Re: Simplifying trig expression

Got a sign wrong in my last reply. Should read
tan(2pi/3-x)=(-root3-t)/(1-root3*t) So cot(2pi/3-x)=(1-root3*t)/(-root3-t)=-(1-root3*t)/(root3+t)= (root3*t-1)/(root3+t)