Finding the sin inverse

Question

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?

Answer 1

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.

Answer 2

+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.

Answer 3

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).

Answer 4

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

$$\implies\dfrac{\sqrt{2-\sqrt3}}2=\dfrac{\sqrt3-1}{2\sqrt2}$$ 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 $x$

Answer 5

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: $$iy=\log\left(ix+\sqrt{1-x^2}\right)$$ $$\arcsin x=-i\log\left(ix+\sqrt{1-x^2}\right)$$