Hello, needmathhelptoujours!

We have: .$\displaystyle \cot2\theta \:=\:\frac{4}{3}\:=\:\frac{adj}{opp}$

$\displaystyle 2\theta$ is in a right triangle with: .$\displaystyle opp = 3,\;adj = 4$ Code:

*
* |
* |
* | 3
* |
* 2θ |
* - - - - - *
4

Pythagorus says: .$\displaystyle hyp = 5$

. . Hence: .$\displaystyle \cos2\theta \:=\:\frac{4}{5}$

We have these two half-angle identities:

. . $\displaystyle \sin\theta \:=\:\sqrt{\frac{1-\cos2\theta}{2}} \qquad\cos\theta \:=\: \sqrt{\frac{1+\cos2\theta}{2}} $

Therefore: .$\displaystyle \begin{Bmatrix}\;\sin\theta &=& \sqrt{\dfrac{1-\frac{4}{5}}{2}} &=& \sqrt{\dfrac{1}{10}} &=& \dfrac{1}{\sqrt{10}}\; \\ \\[-3mm]

\;\cos\theta &=& \sqrt{\dfrac{1+\frac{4}{5}}{2}} &=& \sqrt{\dfrac{9}{10}} &=& \dfrac{3}{\sqrt{10}}\; \end{Bmatrix}$