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

Question

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?

Answer 1

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.

Answer 2

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.