# Finding the sin inverse

Well, so I was finding the value of $\sin 15^{\circ}$. I used the identity $2 \sin x \cos x = \sin 2x$. So solving the quadratic equation and picking the right values simply gave me:

$$\sin 15^{\circ} = \frac {\sqrt {2 - \sqrt {3}}} {2}$$

That was pretty much simple. But I thought what if I was given the value instead, and I was asked to find the angle, whose sine would give me that value.

In short, my question is how to evaluate:

$$\sin^{-1} \left( \frac {\sqrt {2 - \sqrt {3}}} {2} \right)$$

I want to solve it thinking that I do not know the value of $\sin 15$, as if I have been just provided with the problem, and I have no idea what the answer might be. I tried to proceed by converting the $\sin$ to $\tan$ but that seemed to be of no use. Can anybody help?

• $$15=45-30=60-45$$ – lab bhattacharjee Feb 23 '18 at 18:44
• You are doing that because you know that the answer is 15. How is someone supposed to know that just on seeing the problem? – Manish Kundu Feb 23 '18 at 18:48
• Please find my answer – lab bhattacharjee Feb 24 '18 at 10:46

In general it's hard, if not impossible. In this case, it turns out to be doable.

The main idea is that if $\theta = \sin^{-1}\left( \frac{\sqrt{2 - \sqrt 3}}{2} \right)$, then $\sin \theta = \frac{\sqrt{2 - \sqrt 3}}{2}$, and we have an equation we can (hopefully) solve for $\theta$.

Squaring, we get $$\sin^2 \theta = \frac{2 - \sqrt{3}}{4},$$ and multiplying by 4, $$4\sin^2 \theta = 2 - \sqrt{3}.$$

My thoughts are that maybe if I get $\sqrt{3}$ by itself and square, I'll get a nice equation that's polynomial in $\sin \theta$ with integer coefficients. So, we get $\sqrt{3}$ by itself: $$\sqrt{3} = 2 - 4 \sin^2 \theta.$$ But that's interesting: I can factor 2 out... $$\sqrt{3} = 2(1 - 2 \sin^2 \theta),$$

and that $1 - 2 \sin^2 \theta$ looks pretty good, because that's the output from a double angle identity, namely $\cos(2 \theta)$. Now, $$\sqrt{3} = 2 \cos(2 \theta),$$ so $\cos (2\theta) = \frac{\sqrt{3}}{2}$.

This happens when $2 \theta = 30^\circ$ or $2\theta = 330^\circ$.

So, we get one possibility that $\theta = 15^\circ$ (since we've solved an equation by squaring, though, we should make sure it's not extraneous!).

And of course, once we know one angle $\theta$ with a particular sine, we can use symmetry and the unit circle to find all other such angles (that would be $180^\circ - \theta = 180^\circ - 15^\circ = 165^\circ$ in this case, and everything coterminal with the two between $0^\circ$ and $360^\circ$). If we just want the inverse sine of something positive, it's an angle in quadrant I.

• Sidenote, running off to class now, won't be able to respond for a few hours (if it matters) – pjs36 Feb 23 '18 at 19:07
• That was really amazing. Didn't expect it to be solved so simply. – Manish Kundu Feb 23 '18 at 19:23
• @ManishKundu Thanks! I was surprised too; it's a really nice question, and I'm happy you thought to ask it. – pjs36 Feb 24 '18 at 2:26

$$2-\sqrt3=\dfrac{(\sqrt3-1)^2}2$$

$$\implies\dfrac{\sqrt{2-\sqrt3}}2=\dfrac{\sqrt3-1}{2\sqrt2}$$ as $\sqrt3-1>0$

$$=\sin60^\circ\cos45^\circ-\sin45^\circ\cos60^\circ$$

$$=\sin(60-45)^\circ$$

Now we know the principal value of $\sin^{-1}x$ lies in $\in[-90^\circ,90^\circ]$ for real $x$

• You made it simple. That was nice. – Manish Kundu Feb 24 '18 at 15:41

+1 to pjs36's answer, and they did bring up a good point that in this case it was possible to solve. But what if we weren't able to use the double angle identity? Perhaps another way to go would be the use of MacLaurin series.

Given a problem $\sin^{-1}{A} =\theta$, where $\theta$ is in radians and $A \in [-1,1]$, we can first take the $\sin$ of both sides to get $$\sin{\theta} = A$$

The MacLaurin series for $\sin{\theta}$ is $$\sin\theta=\sum_{n=0}^\infty \frac{(-1)^n \theta^{2n+1}}{(2n+1)!} = \theta - \frac{\theta^3}{3!}+\frac{\theta^5}{5!} - \dots =A$$

The farther out you extend the polynomial, the better the estimate is going to be when you solve for $\theta$. However it will be only an estimate when you do it this way, and you would probably need to use a calculator or WolframAlpha or something to solve it as well, which probably defeats the purpose of avoiding just plugging in $\sin^{-1} A$ into the calculator in the first place. This is merely another way you could look at the problem, if you wanted to avoid using inverse trig.

• So I need to solve for $\theta$. Wait, so do I need to use Newton Raphson? – Manish Kundu Feb 23 '18 at 19:49
• Correct, we want to solve for $\theta$. However, this would be equivalent to solving the roots for a large polynomial. Newton Raphson could work, as the derivative of a polynomial is simple to find, but it may take alot of time still to do by hand – WaveX Feb 23 '18 at 19:52
• We would also have the case that we would be estimating an estimate. – WaveX Feb 23 '18 at 19:59
• I see now why solving such problems can be so hard in general. – Manish Kundu Feb 23 '18 at 20:00

For a shortcut evaluation of $x$ such that $\sin^{-1}(x) =k$, assuming that $x$ is "small", you could use the simplest Padé approximant $$\sin^{-1}(x)\approx \frac{x}{1-\frac{x^2}{6}}$$ leading to $$x\approx \frac{\sqrt{6 k^2+9}-3}{k}$$

For your case, where $k= \frac {\sqrt {2 - \sqrt {3}}} {2}$, this would give $$x\approx \sqrt{2+\sqrt{3}} \left(\sqrt{48-6 \sqrt{3}}-6\right)\approx 0.255992$$ in radians (that is to say $\approx 14.6673$ in degrees).

• That was helpful. Thanks. – Manish Kundu Feb 24 '18 at 7:10

Not an answer, just an explicit formula for the $$\arcsin$$ function.

First recall that $$\sin x=\frac{e^{2ix}-1}{2ie^{ix}}$$ So if we want the function $$y=\arcsin x$$ we must solve the following equation for $$y$$: $$x=\frac{e^{2iy}-1}{2ie^{iy}}$$ if we set $$u=e^{iy}$$ this becomes much easier: $$x=\frac{u^2-1}{2iu}$$ $$u^2-1=2iux$$ $$u^2-2iux-1=0$$ The use of the quadratic formula then gives $$u=ix+\sqrt{1-x^2}$$ And using $$\log$$ to denote the complex natural logarithm: $$iy=\log\left(ix+\sqrt{1-x^2}\right)$$ $$\arcsin x=-i\log\left(ix+\sqrt{1-x^2}\right)$$