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?
 A: 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.
A: This sort of problem can be solved by reexpressing the outer function in terms of the function whose inverse appears in the argument. That is, in this case, we need to reexpress the cosine in terms of a tangent. Ignoring sign issues in taking square roots (since the cosine and tangent of the angle of interest are positive), we have
$$\tan\phi=\frac{\sin\phi}{\cos\phi}=\frac{\sqrt{1-\cos^2\phi}}{\cos\phi}\;,$$
$$\cos^2\phi\tan^2\phi=1-\cos^2\phi\;,$$
$$\cos^2\phi(\tan^2+1)=1\;,$$
$$\cos\phi=\frac{1}{\sqrt{\tan^2\phi+1}}\;.$$
Thus
$$
\begin{eqnarray}
\cos(\arctan(3/4))&=&\frac{1}{\sqrt{\tan^2\left(\arctan(3/4)\right)+1}}\\
&=&\frac{1}{\sqrt{(3/4)^2+1}}\\
&=&\frac{1}{\sqrt{(5/4)^2}}\\
&=&\frac{4}{5}\;.
\end{eqnarray}
$$
A: The cosine is defined as the ratio $$\cos{\theta} = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$$ of the sides of right-angle triangle (see image).
Similarly the tangent is $$\tan{\theta} = \frac{\mathrm{opposite}}{\mathrm{adjacent}}$$ 
So $$\theta = \arctan \frac{\mathrm{opposite}}{\mathrm{adjacent}}$$
Since it is a right angle triangle and you know opposite and adjacent sides, you can work out the hypotenuse ($ = \sqrt{op^2 + adj^2}$) and then the cosine.

A: Let $\theta=\arctan(3/4)$.  We want to draw a right-angled triangle $ABC$, with say the right angle at $C$, and the angle $\theta$ at $A$.
So the tangent of the angle at $A$ should be $3/4$. Thus "opposite" divided by "adjacent" should be $3/4$.  This is easy to arrange by making the leg opposite $A$ equal to $3$ and the leg adjacent to $A$ equal to $4$.  
By the Pythagorean Theorem, the hypotenuse is $5$. Now you can read $\cos\theta$ off the picture. It is $4/5$.
