<s>[INST] Trigonometric Identities: Given $\cot(b)=-2$ find $\sin(4b)$ and $\cos(4b)$ 
Given $\cot(b)=-2$ find $\sin(4b)$ and $\cos(4b)$

I found $\sin(4b)$ as $-24/25$ but I always get a very bad answer for $\cos(4b)$
Any hints?
edit:
I made $sin(4b)$ into a expanded form.
$4sin(b)cos^3(b)-4sin^3(b)cos(b)$
And then I made a triangle using $cot(b)=-2$ for information.
I got from that triangle that, $sin(b)=\frac{\sqrt{5}}{5}$ and that $cos(b)=\frac{-2\sqrt{5}}{5}$
And from their I simplified.
But whenever I expand $cos(4b)$ and plug in I get a bad answer, $\frac{353}{25}$ and I do not know if it is correct or wrong.
 [/INST] The result $353/25$ is surely wrong, as a cosine cannot be $>1$.
You're complicating your own life! ;-)
There are standard formulas:
\begin{align}
\sin 2x&=\frac{2\tan x}{1+\tan^2x}=\frac{2\cot x}{\cot^2x+1}
\\[4px]
\cos 2x&=\frac{1-\tan^2x}{1+\tan^2x}=\frac{\cot^2x-1}{\cot^2x+1}
\end{align}
Thus
$$
\sin 2b=\frac{-4}{4+1}=-\frac{4}{5}
$$
and
$$
\cos 2b=\frac{4-1}{4+1}=\frac{3}{5}
$$
Now you know that
\begin{align}
\sin 4b&=2\sin2b\cos2b=-\frac{24}{25} \\[4px]
\cos 4b&=\cos^22b-\sin^22b=…
\end{align}
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