How to solve questions like arcsin(sin 10)?

Question

Ive been reading properties of inverse trigonometrical functions and the questions like:

What will $ \arcsin(\sin 10)$ be equal to?

confuse me a little.

Okay so I understand that $\arcsin(\sin 10)$ cant be directly equal to $10$ as it isn't an element of the principal branch of $\arcsin(\cdot)$ function. But then in the book it is given that

$3\pi-10$ will be the answer and $\sin(3\pi-10)=\sin(2\pi+(\pi-10))=\sin(\pi-10)=\sin10$

Now I understand it but I don't understand from where to get $3\pi-10$,I mean in questions like this what is the first step or the trick to get $3\pi-10$?

Answer 1

I have a shortcut

$\arcsin(\sin(x))$ has a positive slope for

$$ x \in ((4n-1)\dfrac{π}{2},(4n+1)*\dfrac{π}{2})$$ where n is any integer.

If x doesn't fall in this range, the graph has a positive slope.

For a positive slope, $\arcsin(\sin(x)) = kπ + |x|$ and for a negative slope, $\arcsin(\sin(x))= kπ-|x|$.

The value of k, can be determined by approximating the value of x in radians, and ensuring that $kπ \pm |x|$ lies in the range of arcsin function, i.e. $[-π/2,π/2]$.

For your question, $\frac{5π}{2} < 10 < \frac{7π}{2}$.

5 can't be written as 4n-1 for any integer n, so we must have negative slope.

$\arcsin(\sin(10))= kπ - 10$. Only for k=3, kπ-10 lies between $[-π/2,π/2]$.

So $\arcsin(\sin(10))= 3π - 10$.

Also, for $\arccos(\cos(x))$, it has positive slope for $x$ in (even π, odd π) and negative slope for $x$ in (odd π, even π)

arctan(tan(x)) always has a positive slope.

You can also determine:

$\arcsin(\cos(x)) = π/2 - \arccos(\cos(x))$

and $\arccos(\sin(x))= π/2 - \arcsin(\sin(x))$

All these are a consequence of their graphs.

Answer 2

Use the reference angle: subtract $\pi$ enough times from $10$ until you get an angle that is (in absolute value) less than $\pi/2$. So we find that $10-3\pi$ is positive and less than $\pi/2$. Then $\sin(10)$ will be $\pm \sin(10-3\pi)$. To decide the sign, note that $10\mbox{ radians}=(10 \cdot 180/\pi) ^\circ$ is in Quadrant 3, so the sine is negative, and we must have $\sin(10)=-\sin(10-3\pi)=\sin(3\pi-10)$. But now $3\pi-10$ is the domain of the restricted sine function, of which $\arcsin$ is the inverse, so we can just cancel sin with arcsin and get $\arcsin(\sin(10))=\arcsin(\sin(3\pi -10))=3\pi-10$.