3

I want to prove that $\sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}$

Let $y = \sin(\arctan(x))$ and $z=\arctan(x)$. Then, $y = \sin(z)$ $$\frac{\sin(z)}{\cos(z)}=\tan(z)= \frac{y}{\pm \sqrt{1-y^2}}$$ $$\iff \tan(\arctan(x))= \frac{y}{\pm \sqrt{1-y^2}} \\ \iff x^2= \frac{y^2}{1-y^2} $$ $$\therefore y=\pm \frac{x}{\sqrt{1+x^2}}$$ However, I cannot get rid of the negative sign.

Dstarred
  • 2,487
pie
  • 4,192
  • 2
    Nobody can't. As far as I can see, many 'standard' formulae invovling arctan are arbitrary. – ajotatxe Jun 11 '23 at 03:19
  • 2
    If $\theta = \arctan(x) \in [0, \pi/2)$ you can just draw a right triangle with angle $\theta$, opposite side $x$, adjacent side $1$, hypotenuse ...? – Torsten Schoeneberg Jun 11 '23 at 03:26
  • 2
    @ajotatxe: what do you mean? The definitions of arctan and the other inverse trigonometric functions are arbitrary (like every definition) but as soon as a definition is fixed, all formulas can be proved rigorously. AFAIK, the definition of arctan as the inverse function of the restriction of tan on $(-\pi/2,\pi/2)$ is pretty widespread among mathematicians (complex analysts may prefer a definition as a multi-valued function though, but the formulas they prove don't lack rigor, while computer scientists may define of arctan as a 2-variable functions). – Taladris Jun 11 '23 at 04:31
  • 1
    @TorstenSchoeneberg: if $x$ is a real number, $\arctan(x)\in(-\pi/2,\pi/2)$. Of course, the function $\sin(\arctan x)$ is odd, so we can reduce the study of the function to the case $x\ge 0$ (\equivalently $\theta=\arctan(x)\in[0,\pi/2)$). – Taladris Jun 11 '23 at 04:33

5 Answers5

6

The usual convention is that $\arctan x\in (-\pi /2,\pi /2)$ for real $x$.

Let $x=\tan y$ with $y\in (-\pi /2,\pi /2).$ Then $\arctan x=y.$

So on the LHS we have $$\sin (\arctan x)=\sin y.$$ On the RHS we have $$\frac {x}{\sqrt {1+x^2}}=\frac {\tan y}{\sqrt {1+\tan^2 y}}=$$ $$= \frac {\tan y}{\sqrt {1/\cos^2 y}}=$$ $$=\frac {\sin y}{\cos y} \cdot |\cos y|=$$ $$=\sin y$$ because $y\in (-\pi /2,\pi /2)\implies \cos y=|\cos y|.$

DanielWainfleet
  • 57,985
4

Let $\theta = \arctan(x)$. Therefore, $\theta = \arctan(x) \Leftrightarrow x = \tan(\theta)$. We can also write this as $$\frac{x}{1} = \tan(\theta)$$ If we represent this on a triangle and use Pythagoras it is easy to see that $x^2 + 1^2 = c^2 \implies c = \sqrt{x^2 + 1}$, where $c$ is the hypotenuse. Therefore, since sin is opposite over hypotenuse, we have: $$\sin(\theta) = \sin(\arctan(x)) = \frac{x}{\sqrt{x^2 + 1}}$$ As required.

The simplification $\sin(\arctan(x)) = \frac{x}{\sqrt{x^2 + 1}}$ is derived from considering a right triangle or the unit circle, and it holds for all real $x$. This expression will yield a positive value when $x > 0$, and a negative value when $x < 0$, indicating that negative values of $x$ are indeed taken into account in the simplification.

Bumblebee
  • 1,209
4

Let $\theta=\arctan(x)\in\left(-\frac\pi2,\frac\pi2\right)$. Then $\sec(\theta)\gt0$, and $$ \begin{align} \sin(\theta) &=\frac{\tan(\theta)}{\sec(\theta)}\tag{1a}\\ &=\frac{\tan(\theta)}{\sqrt{\tan^2(\theta)+1}}\tag{1b}\\ &=\frac{x}{\sqrt{x^2+1}}\tag{1c} \end{align} $$ Note that $\text{(1b)}$ follows from $\sec(\theta)\gt0$ and $\sec^2(\theta)=\tan^2(\theta)+1$, which is essentially the Pythagorean Theorem.

robjohn
  • 345,667
1

Hint #$1$: When you take the square root of your third line, how many answers do you get?

Hint #$2$: In which quadrants are $x = \tan \theta \Leftrightarrow \theta = \arctan (x)$ positive?

bjcolby15
  • 3,599
-2

Read my comment, please.

I'd write $$y^2=\frac{x^2}{1+x^2}$$ instead.

ajotatxe
  • 65,084
  • Oh, thank you very much for the downvote. A suggestion would even be nicer. – ajotatxe Jun 11 '23 at 03:33
  • this is not me, btw why wolfram alpha suggest that the answer will be positive and also a graphing calculator also suggest that the answer is with positve sign – pie Jun 11 '23 at 03:44
  • I am one of the downvoters. In my opinion, this answer does not answer the question, as the question was about the last step of the computation: choosing the sign in the expression $\pm \frac{x}{\sqrt{1+x^2}}$. – Taladris Jun 11 '23 at 04:37
  • I answered it. You can´t choose the sign. – ajotatxe Jun 11 '23 at 04:48
  • @Taladris I would, in fact, be neutral with this answer. In OP's method, Step 2; $$\tan(\arctan(x))= \frac{y}{\color\red{\pm} \sqrt{1-y^2}}$$ If the working assumes negative, then this should be consistent with the final answer. It should not be "ignored" after squaring in step 3. – Dstarred Jun 11 '23 at 04:48
  • @Dstarred: "the working assumes negative". I don't understand what this means. What "working" – Taladris Jun 11 '23 at 04:53
  • @Taladris as I stated above, OP accounted from negative values in his method. See OP's method in the question. The answer should reflect the method. – Dstarred Jun 11 '23 at 04:56
  • 1
    @ajotatxe: you can. Assuming a standard definition of arctan (cf. Wikipedia), then $\arctan(x)\in(0,\pi/2)$ when $x>0$, so $\sin(\arctan(x))>0$. Since $\frac{x}{\sqrt{1+x^2}}>0$, we have $\sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}$ in that case. Similarly, when $x<0$, both $\sin(\arctan(x))$ and $\frac{x}{\sqrt{1+x^2}}$ are both negative, so $\sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}$ in that case too. (The case $x=0$ is obvious) – Taladris Jun 11 '23 at 04:57
  • @Dstarred: I still don't get your point. First, the OP says: "I want to prove that $\sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}$ (implicitly "for all real numbers" since the LHS and the RHS are both defined on $\mathbb R$). Then they show that "y=\pm \frac{x}{\sqrt{1+x^2}}" which is, up to a sign, the formula they want to prove. They then conclude by "I cannot get [rid] of the negative sign" which clearly shows that they don't know how to finish the proof. I claim that this answer does not finish the proof since it reads to me as "You cannot finish the proof as the given formula is incorrect" – Taladris Jun 11 '23 at 05:03
  • @Taladris In this matter of fact, I do agree with you that OP cannot decide themselves, which domain they are considering. Initially, I was going to comment OP to choose an appropriate domain for $x$, such that his answer is positive. – Dstarred Jun 11 '23 at 05:06
  • @Dstarrted: see also ajotatxe's comment on the question. (I think it should read as "Nobody can" instead of "Nobody can't" but I am not an English native speaker. I think ajotatxe meant "There is no person that can") – Taladris Jun 11 '23 at 05:07
  • 1
    @Dstarred: I still don't understand. Whatever the domain you choose for $x$, as soon as it is a subset of $\mathbb R$, we have $\sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}$. – Taladris Jun 11 '23 at 05:17
  • @Taladris $\displaystyle\sin(\arctan(x)) = \dfrac{x}{\sqrt{1 + x^2}}$ indeed. However, OP inclusion of negative sign changed the perspective of domain. With the plus-minus, OP implied of $-$ for certain values, and hence, changes to domain for $x$. – Dstarred Jun 11 '23 at 05:24
  • 1
    @Dstarred: It seems to me that the OP used the formula $\cos(z)=\pm\sqrt{1-\sin^2(z)}$ (which is true for every $z\in\mathbb R$) and ignore that $z=\arctan(x)\in(-\pi/2,\pi/2)$ (which implies here that $\cos(z)>0$, hence $\cos(z)=\sqrt{1-\sin^2(z)}$). – Taladris Jun 11 '23 at 05:29
  • @Dstarred: The symbol $\pm$ is tricky to use since it can be ambiguous. Are you saying that $f(x)=\pm g(x)$ mean "$f(x)=g(x)$ for some values of $x$ and $f(x)=-g(x)$ for some values of $x$" here? Not "for every $x$, ($f(x)=g(x)$ or $f(x)=-g(x)$)"? For example, do you consider "$a=b \implies a=\pm b$" to be a wrong statement? – Taladris Jun 11 '23 at 05:30