Reference Triangle Issue - Trig Evaluate $\cos(2\cot^{-1}(32/49))$
What I did for this problem was:


*

*Turn $\cot$ into $\tan(49/32)$

*And I am not sure what I do with the $2*$


I tried multiplying $49$ by $2$ and plugging the values into a reference triangle. 
After I had all three sides I just found the $\cos$ of the triangle and used that as the answer but it was incorrect. I also knew i couldn't use any of the double angle formulas because the two is in front of the identity. 
The cot is inverse
Where did I go wrong?

By a well-known formula you have
$$\sin\left(\cot^{-1}x\right)=\frac1{\sqrt{1+x^2}}$$
And, of course, the double angle formula
$$\cos2x=1-2\sin^2x$$
Now your solution is straightforward
$$\begin{align}\cos\left(\cot^{-1}\frac{32}{49}\right)&=1-2\sin^2\left(\cot^{-1}\frac{32}{49}\right)\\[10pt] &=1-2\left(\frac1{\sqrt{1+\frac{32^2}{49^2}}}\right)^2\\[10pt] &=-\frac{1377}{3425}\end{align}$$
