Trigonometric Identities: Given $\cot(b)=-2$ find $\sin(4b)$ and $\cos(4b)$

Question

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.

Answer 1

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}