evaluating compositions of trigonometric functions that do not use standard points or angles from the unit circle I need to find the exact value (so not with a calculator) of trig functions such as the following: cos(arctan(3/4))
I know how to solve these compositions when they use standard points and values from the unit circle but when they use any other points or values I have no idea where to start. My textbook says something about using a triangle but I am not quite understanding the example :/ could anyone lend a hand?

Let $\theta = \arctan(3/4)$. Draw a right triangle with an angle $\theta$ that has that tangent: since $\tan(\theta)$ equals the length of the opposite side divided by the length of the adjacent side, the simplest way to draw such a triangle is to make the opposite side have length 3, and the adjacent side have length 4.
That means that the hypothenuse has length $\sqrt{3^2+4^2} = \sqrt{25} = 5$, by the Pythagorean theorem.
Now you can just read off what $\cos(\theta) = \cos(\arctan(3/4))$ is: since the adjacent side to $\theta$ has length $4$ and the hypothenuse has length $5$, then
$$\cos(\arctan(3/4)) = \cos(\theta) = \frac{\text{adjacent}}{\text{hypothenuse}} = \frac{4}{5}.$$

Alternatively, you can use the basic properties of the angles. Let $\theta$ be an angle with $\tan(\theta) = \frac{3}{4}$. Then
$$\sin^2\theta + \cos^2\theta = 1.$$
Dividing through by $\cos^2\theta$ we get
$$\tan^2\theta + 1  = \frac{1}{\cos^2\theta}.$$
Since $\tan(\theta) = \frac{3}{4}$, then
$$\tan^2\theta + 1 = \left(\frac{3}{4}\right)^2 + 1 = \frac{9}{16}+1 = \frac{25}{16}.$$
So
$$\begin{align*}
\frac{25}{16} &= \frac{1}{\cos^2\theta}\\
\cos^2\theta &= \frac{16}{25}&\quad&\text{(cross-multiplying)}\\
|\cos\theta| &=\sqrt{\frac{16}{25}} = \frac{4}{5}.
\end{align*}$$
Since $-\frac{\pi}{2}\lt \arctan(3/4)\lt \frac{\pi}{2}$, then $\theta$ is in either the first or fourth quadrants, so $\cos\theta$ is positive. Therefore, $\cos\theta = \frac{4}{5}$, same as before.
