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$$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$$

Why can't I figure this one out?? Is it possible to cancel out the arcsins? I know from graphing on my calculator that the answers are $-1, 0$, and $1$, but want help getting there by hand.

Thank you!

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  • $\begingroup$ Sine both sides. $\endgroup$ – Kenny Lau Sep 27 '17 at 10:55
  • $\begingroup$ I tried that but my trig is really rusty - does the sin apply to the whole side or can I individually sin expressions? I mean, sin(arcsinx-sin(x/2))=sin(arcsin((xsqrt3)/2)) so then the right side clearly goes to (xsqrt3)/2 but the left side... what? $\endgroup$ – Emka Sep 27 '17 at 10:56
  • $\begingroup$ @Emka $\sin(A+B)=\sin A\cos B+\sin B\cos A$. You will need model triangle(s). $\endgroup$ – JP McCarthy Sep 27 '17 at 10:57
  • $\begingroup$ By trial, we see some solutions are $x\in \{0,-1,1\}$. $\endgroup$ – samjoe Sep 27 '17 at 11:11
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this makes sense ony if $|x|<1$ \begin{split} &&\arcsin x - \arcsin \dfrac x 2 = \arcsin \dfrac{x\sqrt3}2 \\ &&\sin\left(\arcsin x - \arcsin \dfrac x 2\right) =\sin\arcsin \dfrac{x\sqrt3}2 \\ &&\sin(\arcsin x )\cos(\arcsin \frac x 2)- \cos(\arcsin x )\sin (\arcsin \dfrac x 2) = \frac{x\sqrt3}{2} \\ &&x\sqrt{1-\left(\dfrac x2\right)^2} - \dfrac x2 \sqrt{1-x^2} =\dfrac{x\sqrt3}2 \\ &&x\sqrt{4-x^2} - x \sqrt{1-x^2} = x\sqrt3 \\ && x^2\left(5-2x^2 - 2 \sqrt{(4-x^2)(1-x^2)}\right)= 3x^2 \\ \end{split}

Hence $$x= 0 \qquad or \qquad 5-2x^2 - 2 \sqrt{(4-x^2)(1-x^2) } = 3$$

However, \begin{split} &&5-2x^2 - 2 \sqrt{(4-x^2)(1-x^2) } = 3\\ &\implies & \sqrt{(4-x^2)(1-x^2) } = 1-x^2\\ &\implies & \sqrt{(4-x^2) } =\sqrt{(1-x^2) } \qquad or \qquad 1-x^2 = 0 \end{split} But $ \sqrt{(4-x^2) } =\sqrt{(1-x^2) }$ is impossible

Conclusion $x\in \{-1,0,1\}$ is the set of solutions

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  • $\begingroup$ Thank you so much, I really appreciate it! I've been bothered by this problem for a week, but now it's clear. $\endgroup$ – Emka Sep 27 '17 at 13:20
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Use $\sin(A-B)$ identity and the fact that $f(f^{-1}(x)) = x$:

$$x\sqrt{1-\frac{x^2}{4}}-\frac{x}{2}\sqrt{1-x^2} = \frac{x\sqrt 3}{2}$$

So $x=0$ is one solution. Now

$$\sqrt{4-x^2}-\sqrt{1-x^2} = \sqrt{3}$$

Now let $1-x^2 = t$, so that our equation becomes:

$$\sqrt{3+t}-\sqrt{t} = \sqrt{3}$$

Upon squaring, we get:

$$3+t+t - 2\sqrt{(3+t)t} = 3 \\ t^2 = (3+t)(t)$$

From here, we get $t = 0$. Thus $1-x^2 = 0$, so that $x = \pm 1$

So we have three solutions, $x = -1, 0, 1$, which satisfy the original equation.

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