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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.

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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 $O = x$.

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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 $$\sin(\tan^{-1}(2)) = \frac 2{\sqrt{1 + 2^2}} = \frac2{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$

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  • $\begingroup$ Thank you so much for this answer, this was explained just how I needed it. $\endgroup$ Commented Mar 27, 2018 at 5:28
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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?

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    $\begingroup$ This may sound stupid, but how exactly did you translate $$(\tan^{-1}(2))$$ into the triangle with side lengths 1 and 2. $\endgroup$ Commented Mar 27, 2018 at 5:20
  • $\begingroup$ In a right triangle, the value of the tangent function is defined as the ratio between the opposite and adjacent sides. Therefore, if their ratio is two, then $\tan{\theta} = 2$, where $\theta$ is the angle that is opposite the side with length $2$. $\endgroup$
    – Matti P.
    Commented Mar 27, 2018 at 5:38
  • $\begingroup$ @KrishnanshuGupta It doesn't have to be sides of lengths $1$ and $2$. But it does have to be any right triangle whose legs have a ratio of $1:2$. $\endgroup$ Commented Mar 27, 2018 at 15:23
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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 $\sin x > 0$

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  • $\begingroup$ Totally confused about this one, I started inverse functions today, and all these functions just blow my mind even though I've dealt with them without inverses. Can you provide some explanation of how you got to all those steps. Thanks so much. $\endgroup$ Commented Mar 27, 2018 at 5:25
  • $\begingroup$ When dealing with inverses, are we no longer looking at the constraints of the unit circle? Or is it still an angle on the unit circle? Thanks $\endgroup$ Commented Mar 27, 2018 at 5:26
  • $\begingroup$ We define $x = \arctan 2$ so that $\tan x = 2$. The inverse tangent function only has a range of $[-\pi/2, \pi/2]$ $\endgroup$
    – Dylan
    Commented Mar 27, 2018 at 5:43
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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 $\sin \theta = \frac{2}{\sqrt5}=\frac{2\sqrt5}{5}$

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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 $\sin t=\frac {2}{\sqrt 5\;}.$

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  • $\begingroup$ Given any one of $\sin t. \cos t , \tan t,$ you can determine the absolute values of the other two by the formulas $\sin^2 t+\cos^2 t=1$ and $\tan t=\sin t/\cos t.$ $\endgroup$ Commented Mar 27, 2018 at 5:55

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