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Find all real numbers $x$ such that $$\sin(\arccos(\tan(\arcsin x))))=x.$$

Using Wolfram|Alpha we get $$\sin(\arccos(\tan(\arcsin x))))=\sqrt{\dfrac{2x^2-1}{x^2-1}}$$

Why? Because some case such $\arcsin{x}=\arccos{(1-x^2)}$ or $\pi-\arccos{(\sqrt{1-x^2})}.$

I think the answer is $x=\dfrac{\sqrt{5}\pm 1}{2}.$ Am I right?

math110
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4 Answers4

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Let $\arcsin x=y\,$, therfore $\sin y=x$ and $\cos y=+\sqrt{1-x^2}\,$ as $\,-\dfrac\pi2\le y\le\dfrac\pi2$ we have $$\tan(\arcsin x)=\tan(y)=\dfrac x{\sqrt{1-x^2}}$$ thus $$x=\sin\left(\arccos\dfrac x{\sqrt{1-x^2}}\right)$$

set $$\arccos\left(\dfrac x{\sqrt{1-x^2}}\right)=u\,\quad ,\quad0\le u\le\pi$$

we can write $$\sin u=+\sqrt{1-\dfrac{x^2}{1-x^2}}=\sqrt{\dfrac{1-2x^2}{1-x^2}}$$

So, we have $$x=\sqrt{\dfrac{1-2x^2}{1-x^2}}$$ which is $\ge0$

Squaring we get $$x^2(1-x^2)=1-2x^2\iff x^4-3x^2+1=0$$

Solve for $x^2$

Behrouz Maleki
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lab bhattacharjee
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1

Let's try dismantling $\sin(\arccos(\tan(\arcsin x)))$.

Note that $\sin(\arccos t)\ge0$, so we can apply $$ \sin(\arccos t)=\sqrt{1-(\cos(\arccos t))^2}=\sqrt{1-t^2} $$

Thus we have a first reduction $$ \sqrt{1-(\tan(\arcsin x))^2}=x $$ which tells you that $x\ge0$.

Now let's tackle $\tan(\arcsin x)$ over the interval $[0,1)$. We have $$ \tan(\arcsin x)= \frac{\sin\arcsin x}{\cos\arcsin x}=\frac{x}{\cos\arcsin x} $$ Again, $\cos(\arcsin x)\ge0$, so we have $$ \cos(\arcsin x)=\sqrt{1-(\sin(\arcsin x))^2}=\sqrt{1-x^2} $$

Thus the equation is $$ \sqrt{1-\biggl(\frac{x}{\sqrt{1-x^2}}\biggr)^2}=x $$ or $$ 1-\frac{x^2}{1-x^2}=x^2 $$ Recall that only solutions in the interval $[0,1)$ can be accepted.

egreg
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$$\sin(\arccos(\tan(\arcsin x))))=x$$ implies

$$\tan(\arcsin x))=\cos(\arcsin(x))=\pm\sqrt{1-x^2}$$

and in general

$$\tan(\arcsin x))=\pm\frac x{\sqrt{1-x^2}}.$$

Then

$$1-x^2=\pm x.$$

As $|x|\le1$, two candidates are $$\pm\frac{\sqrt5-1}2,$$ but only the positive sign holds. (Denoting the quadrants in Roman numbers, $+\to I\to+\to I\to+$ or $-\to IV\to-\to II\to+$.)

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    $\tan(\arcsin x)=\pm\sqrt{1-x^2}$ seems incorrect – Mc Cheng Jul 03 '16 at 14:55
  • @McCheng: no, you apply $\cos(\arcsin())$ to the two members of the original equation. It is clear that this is not an identity, as the $\tan$ would magically turn to a $\cos$; the next line is an identity. The upvotes to this remark are undue. –  Jul 03 '16 at 15:27
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$$\arccos(\tan(\arcsin x)))=\arcsin x=\text{sign}(x)\cdot\arccos\sqrt{1-x^2}$$

If $x\ge0,$ $$\tan(\arcsin x))=\sqrt{1-x^2}$$

If $\arcsin x=y,\sin y= x,0\le x\le\dfrac\pi2\implies\cos y=+\sqrt{1-x^2}$

$\sqrt{1-x^2}=\tan(y)=\dfrac x{\sqrt{1-x^2}}$

$\iff1-x^2=x$

If $x<0,$

using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?,

$$\arccos(\tan(\arcsin x)))=-\arccos\sqrt{1-x^2}=\pi-\arccos(-\sqrt{1-x^2})$$

$$\arccos(-\sqrt{1-x^2})=\pi-\arccos(\tan(\arcsin x)))=\arccos(-\tan(\arcsin x)))$$

$$\sqrt{1-x^2}=\tan(\arcsin x)$$

Can you take it from here?

Community
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lab bhattacharjee
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