How to find $\sin^{-1}(\sin{100})$

To solve such questions we were told to infer the graph of $\sin^{-1}(\sin{x})$ but this seems too big of a number.

Is there any shorter way to solve such questions?

2 Answers

Remember that $$\sin^{-1}(\sin x)=x+2k\pi \;\text{ or }\; (\pi-x)+2k\pi$$ for some $k\in \mathbb Z$.

So, $$\sin^{-1}(\sin 100)=100+2k\pi \;\text{ or }\; (\pi-100)+2k\pi$$ for some $k\in \mathbb Z$.

Now, fix the value of $k$ such that either $$-\frac \pi 2 \leq 100+2k\pi \leq \frac \pi 2$$ or $$-\frac \pi 2 \leq (\pi-100)+2k\pi \leq \frac \pi 2$$ holds, since the range of $\sin^{-1}x$ is defined to be $\bigg [-\dfrac {\pi} {2}, \dfrac \pi 2\bigg]$

This gives you the answer.

Sorry for the late answer. We will try to solve this easily by two methods. The above answers also try to convey the same thing.

1. We know that $f(x)=\sin^{-1}(\sin{x})=x$ when $x\in\bigg[-\dfrac{\pi}{2},\dfrac{\pi}{2}\bigg]$.
   So now our main aim is to somehow convert $100$ (radians) into this range by using the property $\sin{x}=\pm \sin({n\pi-x})$ depending on the value of $n$.
   The closest multiple of $\pi$ to $100$ is $32\pi\approx 100.531$. Now rewriting $\sin{100}=\sin(-({32\pi-100}))$ where '$-$' is added because $32\pi-100$ lies in the $4^{\text{th}}$ quadrant. So now we have $f(x)=\sin^{-1}(\sin({100-32\pi}))=100-32\pi$ as it clearly lies in the mentioned interval.
2. We can derive a general pattern for this curve from the graph. Also, $f'(x)=\pm1$ for any interval of $x$.

$$ f(x)= \begin{cases} x, &\text{when} & x \in \bigg[-\dfrac{\pi}{2},\dfrac{\pi}{2}\bigg]\\ \pi-x, &\text{when} & x \in \bigg[\dfrac{\pi}{2},\dfrac{3\pi}{2}\bigg]\\ x-2\pi, &\text{when} & x \in \bigg[\dfrac{3\pi}{2},\dfrac{5\pi}{2}\bigg] &\text{so on} \dots \\ \end{cases} $$ So writing in a general way, $$ f(x)= \begin{cases} \text{(odd multiple of $\pi$)}-x\\ x-\text{(even multiple of $\pi$)} \end{cases} $$ You can choose a multiple of $\pi$ (even/odd) above ($2$) in such a way that $f(x)$ should lie in the required interval i.e, $x\in \bigg[-\dfrac{\pi}{2},\dfrac{\pi}{2}\bigg]$ by checking both possibilities and choosing that one which gives answer in the required interval.
To sum up, you just have to find the closest multiple of $\pi$ to the given number and then can you can apply any of these above methods. You'll get a hang of it once you practice some problems.