Solve the following equation $$2\arccos(x)-\arcsin(2x\sqrt{1-x^2})=0.$$
I tried to solve this equation, but I don't know how can I change $\arcsin$ to $\arccos$ to solve the equation.
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Robert Z
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I tried to use: arcsin x = ( pi/2 - arccos x) However, it didn't work because my arcsin would get a coefficient of 2. – Nariman Zendehrooh Sep 15 '17 at 05:15
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Note that $x\in [-1,1]$. Let $x=\cos(t)$ with $t\in [0,\pi]$ then $\sin(t)\geq 0$ and $$2\arccos(\cos(t))=\arcsin(2\cos(t)\sqrt{1-\cos^2(t)})=\arcsin(2\cos(t)\sin(t))=\arcsin(\sin(2t)).$$ Note for $t\in [0,\pi]$, $\arccos(\cos(t))=t$ and (see HERE) $$\arcsin(\sin(2t))=\begin{cases} 2t &\mbox{if $t\in [0,\pi/4]$}\\ \pi-2t &\mbox {if $t\in [\pi/4,3\pi/4]$}\\ -2\pi+2t &\mbox {if $t\in [3\pi/4,\pi]$} \end{cases}$$ Can you take it from here?
Robert Z
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Not really! How did you get sin(t) from (1-x^(1/2))^(1/2)? And what can I do with 2t? – Nariman Zendehrooh Sep 15 '17 at 05:24
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Now it makes sense! So, then if I take sin from both sides of the equation I will get sin (2t)= sin (2t) which does not make sense again... – Nariman Zendehrooh Sep 15 '17 at 05:41
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$\sin$ is not an bijective function in $[0,\pi]$. therefore $\sin (2t)= \sin (2t)$ is not equivalent to the equation $2t=\arcsin(\sin(2t))$. Did you take a look to the link that I gave you? – Robert Z Sep 15 '17 at 05:45
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for t from 0 to pi/4 I would get infinite solutions. and for pi/4 to 3pi/4 I would get x= cos(pi/4). Is this the answer? – Nariman Zendehrooh Sep 16 '17 at 00:28
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1@NarimanZendehrooh Yes, therefore the solutions are $x\in [\cos(\pi/4),\cos(1)]=[1/\sqrt{2},1]$. – Robert Z Sep 16 '17 at 06:29
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Hmmm what about the following approach ( there is a fair bit of trig )
Take the cosine of both sides to get
$$ \cos\left(\underbrace{2 \cos^{-1}(x)}_{\text{=Part 1}} - \underbrace{\sin^{-1}\left(2x\sqrt{1-x^2}\right)}_{\text{=Part 2}} \right) = 1. \tag{1} \label{eqn:main} $$
Look at each part individually.
$$ \cos(2 \cos^{-1}(x)) = \cos(2\theta), $$ where $\theta \equiv \cos^{-1}(x)$ and hence $x = \cos(\theta)$. Then use the double angle formula
$$\cos(2\theta) = \cos^2 (\theta) - \sin^2(\theta) \\ = 2\cos^2(\theta) - 1. $$
Then using what we have defined above for $\theta$ and $x$ we may deduce that
$$ cos(2 \theta) = 2x^2 -1 .$$
Okay first part done. Second part is similar. We have
$$ \cos\left(\sin^{-1}\left(2x \sqrt{1 - x^2}\right) \right).$$
Let
$$ \sin^{-1}\left(t\right) \equiv \Theta, $$
where $t \equiv2x \sqrt{1 - x^2}.$ Then
$$ \sin^2 (\Theta) = t^2 \Rightarrow 1- \cos^2(\Theta) = t^2,$$
Or in other words
$$ \Theta = \cos^{-1} \left( \sqrt{1 - t^2} \right).$$
Finally, substituting this back in to the original expression
$$\cos\left(\sin^{-1}\left(2x \sqrt{1 - x^2}\right) \right) = \cos\left( \cos^{-1}(\Theta) \right).$$
So the full equation now reads
$$ 2x^2 - 1 - \sqrt{1 - t^2} = 1,$$
which, I believe after substituting in the expression for $t$ results with solution $ x = \pm 1$.
Rumplestillskin
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Let $\arccos x=t\implies0\le t\le\pi,x=\cos t,\sin t=+\sqrt{1-x^2}$
As $-\dfrac\pi2\le\arcsin(2x\sqrt{1-x^2})\le\dfrac\pi2,$
we have $$-\dfrac\pi2\le2t\le\dfrac\pi2$$
$$\arcsin(2x\sqrt{1-x^2})=\arcsin(\sin2t)$$
$$=\begin{cases} 2t &\mbox{if } -\dfrac\pi2\le2t\le\dfrac\pi2\\ \pi-2t & \mbox{if }-\dfrac\pi2\le\pi-2t\le\dfrac\pi2\iff\dfrac\pi2\le 2t\le\dfrac{3\pi}2 \\2\pi+2t & \mbox{if }-\dfrac\pi2\le2\pi+2t\le\dfrac\pi2\iff ? \\ -2\pi+2t& \mbox{if }-\dfrac\pi2\le-2\pi+2t\le\dfrac\pi2\iff ?\end{cases} $$
But we already have $$-\dfrac\pi2\le2t\le\dfrac\pi2$$
lab bhattacharjee
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