I just need help to finetune this solution. Only having a correct answer is not sufficient to comb home 5/5 points on a problem like this. Thoughts on improvements on stringency? Is there any logical fallacy or ambiguity? Any input is very welcome!

Solution: The domain for $\arctan{x}$ is $\mathbb{R},$ thus the function $\sin{(\arctan{x})}$ is defined over the entire $\mathbb{R}$. However, the restricting factor comes from the fact that the domain for $\arcsin{x}$ is $[-1,1]$ which implies that the domain for $\tan{(\arcsin{x})}$ is $[-\arctan{1},\arctan{1}]=[-\frac{\pi}{4},\frac{\pi}{4}].$ We can thus conclude that if there exists a solution $x$ to the above equation, then $x\in[-\frac{\pi}{4},\frac{\pi}{4}]$.

First I note that $$\sin{b}=\pm\frac{\tan{b}}{\sqrt{\tan^2{b}+1}}\Rightarrow\sin^2{(\arctan{x})}=\left(\pm\frac{\tan{(\arctan{x})}}{\sqrt{\tan^2{(\arctan{x})}+1}}\right)^2=\frac{x^2}{x^2+1}.$$

Secondly; $$\tan{c}=\pm\frac{\sin{c}}{\sqrt{1-\sin^2{c}}}\Rightarrow \tan^2{\arcsin{x}}=\left(\pm\frac{\sin{(\arcsin{x})}}{\sqrt{1-\sin^2{(\arcsin{x})}}}\right)^2=\frac{x^2}{1-x^2}.$$

So the equation becomes $$\frac{x^2+1-1+x^2}{x^2}=\frac{2x^2}{x^2} =4x^2\Longleftrightarrow x=\pm\frac{1}{\sqrt{2}}.$$

What's left to do now is to show that these $x:$s both lie in the desired interval. So it boils down to showing that $\frac{1}{\sqrt{2}}\leq\frac{\pi}{4}$. This can be done by multiplying the inequality by $\sqrt{2}/\sqrt{2}$: $$\frac{\sqrt{2}}{4}\leq \frac{\pi}{4}\Longleftrightarrow \sqrt{2}\leq \pi,$$

which clearly is true.

  • $\begingroup$ I know the result is correct, but for only a correct answer I'd get 1 point of 5. The rest are obtained by a sound mathematical way and a nice mathematical language. $\endgroup$
    – Parseval
    Sep 18, 2017 at 20:22
  • 1
    $\begingroup$ Domain for tan(arcsinx) is (-1,1) $\endgroup$ Sep 18, 2017 at 20:32

2 Answers 2


The allowed values for $x$ are in $[-1,1]$, because of $\arcsin x$, but $0$ should also be excluded. Also $-1$ and $1$ must be excluded because of $\tan\arcsin x$.

(Note: your $x\in[-\pi/4,\pi/4]$ is wrong and the probable cause for the low grade.)

The equation remains the same if we change $x$ into $-x$, so we can limit ourselves to $x\in(0,1)$. For every root we find, also its negative will be a root.

If $y=\sin\arctan x$, then $\arctan x=\arcsin y$ and so $x=\tan\arcsin y$ so $$ x^2=\frac{\sin^2\arcsin y}{1-\sin^2\arcsin y}=\frac{y^2}{1-y^2} $$ whence $$ y^2=\frac{x^2}{1+x^2} $$ If $z=\tan\arcsin x$, then $$ z^2=\frac{\sin^2\arcsin x}{1-\sin^2\arcsin x}=\frac{x^2}{1-x^2} $$ So your equation $\frac{1}{y^2}-\frac{1}{z^2}=4x^2$ becomes $$ \frac{1+x^2}{x^2}-\frac{1-x^2}{x^2}=4x^2 $$ that is, $$ 2x^2=4x^4 $$ and, since $x\ne0$, $x^2=\frac{1}{2}$. The only positive root is $$ x=\frac{1}{\sqrt{2}} $$ which indeed belongs to $(0,1)$. Thus the solutions are $$ \frac{1}{\sqrt{2}}\qquad\text{and}\qquad{-}\frac{1}{\sqrt{2}} $$

  • $\begingroup$ Yes, ofcourse! arctan is undefined at $-\pi/2$, $0$ and $\pi/2$. So the domain is $(-1,0)\cup(0,1)$, if I'm not mistaken? Which in turn means that the solutions to the equation $x\in[\pi/4,\pi/4] / {0}$? $\endgroup$
    – Parseval
    Sep 18, 2017 at 20:53
  • $\begingroup$ @Parseval No: $x\in(-1,0)\cup(0,1)$. Why should you take the arctangent? $\endgroup$
    – egreg
    Sep 18, 2017 at 20:54
  • $\begingroup$ Because it's 22:57 and I've been staring at math for 18 hours. This is the kind of blind mistakes you do in such a condition. Thank you very much for clearing up everything, I really appreciate it mate! $\endgroup$
    – Parseval
    Sep 18, 2017 at 20:57

Taking derivative of LHS we have $$\frac{d}{dx}\left(\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}\right)=0$$ Therefore $$\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}=\text{ constant }$$ Calculation for $x=\dfrac{1}{\sqrt3}$ gives $$\frac{1}{\sin^2{(\arctan{x})}}-\frac{1}{\tan^2{(\arcsin{x})}}=2$$ Now the points of the parabola $y=4x^2$ with $2=4x^2$ are $\color{red}{x=\pm\dfrac{1}{\sqrt2}}$


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