Inverse Tangent Function I recently started working with inverses of trig functions in a textbook, when this problem stumped me: $$\sin(\tan^{-1}(2))$$ This problem is easily solvable using a calculator, but I am wondering how one would do it without a scientific calculator. With a calculator, the answer is $0.894427$, but for some reason, the textbook displays the answer as $\dfrac{2\sqrt5}5$, which equals the same thing, but I would like to know how they came about the answer in that form. Thanks for any suggestions.
 A: Let us look at a general trigonometric function of the form $$\sin(\tan^{-1}(x))\tag{1}$$
Let $$\theta = \tan^{-1}(x)\tag{2}$$ Substitute $(2)$ in $(1)$ to get
$$\sin(\theta)\tag{3}$$
Let's go back to $(2)$.
$$\theta = \tan^{-1}(x)\implies\tan(\theta) = x = \frac x1\tag{4}$$
Imagine the triangle below having length $A = 1$ and height $O = x$.

Using the Pythagorean Theorem, we can deduce the length $H = \sqrt{1^2 + x^2} = \sqrt{1 + x^2}$.
Recall that $$\sin(\theta) = \frac OH = \frac {x}{\sqrt{1 + x^2}}$$
Great! We have an expression for $\sin(\theta)$. Let's convert it to the original form. $$\sin(\theta) = \sin(\tan^{-1}(x)) = \frac {x}{\sqrt{1 + x^2}}$$
Plug in $x = 2$ to get
$$\sin(\tan^{-1}(2)) = \frac 2{\sqrt{1 + 2^2}} = \frac2{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$
A: The value we want to solve is 
$$
x = \sin{\left( \tan^{-1}{2} \right)} \Leftrightarrow \sin^{-1}x = \tan^{-1}2
$$
Draw a right triangle with side lengths $1$ and $2$. Now the tangent of one of the angles is the correct angle. Can you now figure out the sine of the angle?
A: Let $\tan x = 2$, then we're looking for
$$ y = \sin x = \tan x \cos x = \frac{\tan x}{\sqrt{\tan^2 x + 1}} = \dots $$
Note that the angle $0 < x < \pi/2$ is unique, due to the range of the arctangent function. Therefore $\cos x > 0$ and $\sin x > 0$
A: Draw a right angle triangle.
Pick an angle that is not the right angle, call it $\theta$.  label the opposite side as having length $2$, and the adjacent side having length $1$, hence fulfilling $\tan \theta = 2$ and $\theta = \tan^{-1}(2)$.
By pythagoras theorem, we have the hypothenus as $\sqrt{1^2+2^2}=\sqrt{5}$.
Hence $\sin \theta = \frac{2}{\sqrt5}=\frac{2\sqrt5}{5}$
A: When $\cos t \ne 0 $ we have $$1+\tan^2 t=1+\frac {\sin^2t}{\cos^2t}=\frac {\cos^2t+\sin^2t}{\cos^2t}=\frac {1}{\cos^2t}.$$  $$ \text {Therefore }\quad \cos^2t =\frac {1}{1+\tan^2t}$$  $$\text {and also }\quad  \sin^2 t=1-\cos^2t=1-\frac {1}{1+\tan^2 t}=\frac {\tan^2 t}{1+\tan^2 t}.$$   Now $\tan^{-1}2$ is the unique $t\in (0,\pi/2,\pi/2)$ such that $\tan t =2.$  So we have  $$\sin^2 (\tan^{-1}2)=\sin^2 t=\frac {\tan^2t}{1+\tan^2 t}=\frac {4}{5}.$$ Therefore $|\sin t|=\frac {2}{\sqrt 5\;}.$  And since $(t\in (-\pi/2,\pi/2\land \tan t>0)\implies t\in (0,\pi/2)\implies \sin t>0,$ we have $\sin t >0.$ 
Therefore $\sin t=\frac {2}{\sqrt 5\;}.$
