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Given an equation: $$\sin^{-1}(2x) + \sin^{-1}(3x) = \frac \pi 4$$

How do I find $x$?

I tried solving by differentiating both sides, but I get $x=0$.

How do you solve it, purely using trigonometric techniques?

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  • $\begingroup$ Possibly useful - math.stackexchange.com/questions/672575/… $\endgroup$ – Eevee Trainer Jan 8 at 2:50
  • $\begingroup$ But I don't know x^2+y^2 $\endgroup$ – user5722540 Jan 8 at 2:52
  • $\begingroup$ I don't see how anyone can solve this in under two minutes. Are you sure? If so, there must be a trick. Is this a multiple choice question? $\endgroup$ – D.B. Jan 8 at 4:06
  • $\begingroup$ Online open ended answer. Deadline of 150 seconds. $\endgroup$ – user5722540 Jan 8 at 4:09
  • $\begingroup$ Well, it takes Maple less than a second to get ${\frac {\sqrt {2522-1164\,\sqrt {2}}}{194}}$ in response to solve(expand(sin(arcsin(2*x)+arcsin(3*x))=sin(Pi/4)));. Typing the question and answer takes longer... $\endgroup$ – Robert Israel Jan 8 at 4:23
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Or this way using

  • $\cos(a+b) = \cos a \cos b - \sin a \sin b$
  • $\cos a = \sqrt{1-\sin^2 a}$

\begin{eqnarray*} \sin^{-1}(2x) + \sin^{-1}(3x) & = & \frac \pi 4 \\ \sqrt{1-4x^2}\sqrt{1-9x^2} - 6x^2 & = & \frac{\sqrt{2}}{2} \\ (1-4x^2)(1-9x^2) &=& \left( \frac{\sqrt{2}}{2} +6x^2 \right)^2 \\ \frac{1}{2} & = & (6\sqrt{2}+13)x^2 \\ \end{eqnarray*} The positive solution gives: $$\boxed{x = \frac{1}{\sqrt{2(6\sqrt{2}+13)}}}$$

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  • $\begingroup$ The answer is not matching @clathratus solution. Please confirm. $\endgroup$ – user5722540 Jan 8 at 7:48
  • $\begingroup$ Just plug it in. It gives the correct result. Unfortunately I cannot paste the complete WA-link here. $\endgroup$ – trancelocation Jan 8 at 7:50
  • $\begingroup$ @user5722540 : I have just checked with WA that both expressions are equivalent. Happens quite often that nested radical expressions have quite differently looking ways to be expressed. $\endgroup$ – trancelocation Jan 8 at 7:55
  • $\begingroup$ Got that. Thanks. $\endgroup$ – user5722540 Jan 8 at 8:05
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There is a useful identity that we can use in this case:

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

From here we can substitute:

$\arcsin{(2x\sqrt{1-9x^2}+3x\sqrt{1-4x^2})}=\frac{\pi}{4}$

We are then left with:

$2x\sqrt{1-9x^2}+3x\sqrt{1-4x^2}=\sin{\frac{\pi}{4}}$

From here, you can solve for $x$.

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  • $\begingroup$ When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes? $\endgroup$ – user5722540 Jan 8 at 3:18
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Let $\theta_1 = \sin^{-1}(2x)$,$\theta_2 = \sin^{-1}(3x)$. Then, $$\sin(\theta_1+\theta_2) = \sin(\theta_1)\cos(\theta_2)+\sin(\theta_2)\cos(\theta_1) = \sin(\pi/4) = 1/\sqrt{2}$$ $$2x\sqrt{1-9x^2}+3x\sqrt{1-4x^2} = \frac{1}{\sqrt{2}}.$$ But I don't know how you would solve this last equation.

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  • $\begingroup$ When I tried solving the equation, I got x=root(+-(root(17)+5)*10)/20. However, on again solving it, I get a completely different answer. Is the answer actually going to be this complicated, given that it is expected to be solved in under 2minutes? $\endgroup$ – user5722540 Jan 8 at 3:19
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I'm gonna derive the general function for $\arcsin x$ then go from there.

Recall that $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$ So if $y= \arcsin x$, then $$2ix=e^{iy}-e^{-iy}$$ Letting $u=e^{iy}$, we have $$2ix=\frac{u^2-1}{u}$$ $$u^2-2ixu-1=0$$ Use the quadratic formula to find that $$u=ix+\sqrt{1-x^2}$$ Thus $$e^{iy}=ix+\sqrt{1-x^2}$$ $$iy=\ln\big[ix+\sqrt{1-x^2}\big]$$ $$\arcsin x=-i\ln\big[ix+\sqrt{1-x^2}\big]$$ So we look at your equation: $$\arcsin 2x+\arcsin3x=\frac\pi4$$ $$-i\ln\big[2ix+\sqrt{1-4x^2}\big]-i\ln\big[3ix+\sqrt{1-9x^2}\big]=\frac\pi4$$ $$\ln\big[2ix+\sqrt{1-4x^2}\big]+\ln\big[3ix+\sqrt{1-9x^2}\big]=\frac{i\pi}4$$ Using the property $\ln(ab)=\ln a+\ln b$ we see that $$\ln\bigg[\big(2ix+\sqrt{1-4x^2}\big)\big(3ix+\sqrt{1-9x^2}\big)\bigg]=\frac{i\pi}4$$ Taking $\exp$ on both sides, $$\big(2ix+\sqrt{1-4x^2}\big)\big(3ix+\sqrt{1-9x^2}\big)=e^{i\pi/4}$$ Use the formula $e^{i\theta}=\cos\theta+i\sin\theta$ to see that $$\big(2ix+\sqrt{1-4x^2}\big)\big(3ix+\sqrt{1-9x^2}\big)=\frac{1+i}{\sqrt2}$$ and at this point I used Wolfram|Alpha to see that $$x=\sqrt{\frac{13}{194}-\frac{3\sqrt2}{97}}$$ I will update my answer once I figure out how this result is found


Edit:

Expanding out the product on the Left hand side, then multiplying both sides by $-i$, we have $$x\bigg(2\sqrt{1-9x^2}+3\sqrt{1-4x^2}\bigg)+6ix^2-i\sqrt{(1-4x^2)(1-9x^2)}=\frac{1-i}{\sqrt{2}}$$ We set the real parts of each side equal to eachother: $$x\bigg(2\sqrt{1-9x^2}+3\sqrt{1-4x^2}\bigg)=\frac1{\sqrt2}$$ Which @ClaudeLeibovici showed reduced to $$97y^2-13y+\frac14=0$$ with $y=x^2$. Using the quadratic formula, we see that $$y=\frac{13+\sqrt{72}}{194}$$ which reduces to $$y=\frac{13}{194}+\frac{3\sqrt2}{97}$$ Taking $\sqrt{\cdot}$ on both sides, $$x=\sqrt{\frac{13}{194}-\frac{3\sqrt2}{97}}$$

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  • $\begingroup$ Is there a workaround to approach solution under 150seconds? $\endgroup$ – user5722540 Jan 8 at 7:49
  • $\begingroup$ @user5722540 Yes. This is more of an alternative approach using cool complex algebra instead of trig $\endgroup$ – clathratus Jan 8 at 17:54
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We need $-1\le3x\le1$

But if $x\le0,$ the left hand side $\le0$

Now $3x=\sin(\pi/4-\arcsin(2x))$

$3\sqrt2x=\sqrt{1-(2x)^2}-2x$

$\sqrt{1-4x^2}=x(3\sqrt2+2)$

Square both sides

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Start with $$2x\sqrt{1-9x^2}+3x\sqrt{1-4x^2} = \frac{1}{\sqrt{2}}$$ and square both sides to get $$-72 x^4+13x^2+12 \sqrt{1-9 x^2} \sqrt{1-4 x^2} x^2=\frac 12$$ that is to say $$72 x^4-13x^2+\frac 12=12 \sqrt{1-9 x^2} \sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make $$72 y^2-13y+\frac 12=12 y\sqrt{1-9 y} \sqrt{1-4 y} $$ Square again, expand and simplify to get $$97 y^2-13 y+\frac{1}{4}=0$$ which is simple.

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