Find the principle value of $\arcsin(\sin23)$ and $\arcsin(\sin24),$ where both $23$ and $24$ are in radians.
Find the value of $\arcsin(\sin23)$ and $\arcsin(\sin24),$ where both $23$ and $24$ are in radian.
I tried many times to evaluate $\arcsin(\sin23)$ and $\arcsin(\sin24)$ in the same way as the method given in this link, but I am unable to follow his methods for $\arcsin(\sin23)$ and $\arcsin(\sin24).$
Please,help me to evaluate $\arcsin(\sin23)$ and $\arcsin(\sin24),$ using the method mentioned in the link.
Thank you!
$\endgroup$ 71 Answer
$\begingroup$This is a long-winded comment.
This is not an answer.
I am confused by your posting.
Suppose that $N$ is some (known, fixed) real number and you let $t = \sin(N)$. Then, per the definition of the arcsin function, you are looking for some value $\theta$ such that $-(\pi/2) \leq \theta \leq (\pi/2)$ and such that $\sin(\theta) = t.$
You know that the sine function has a period of $(2\pi).$ So, the first thing to do is to compute
$$M = N - \left\{ ~2\pi \times \left\lfloor \frac{N}{2\pi}\right\rfloor ~\right\}.$$
This implies that
- $\sin(M) = \sin(N)$
- and that $0 \leq M < 2\pi.$
Then, it is required that
- $\sin(\theta) = \sin(M)$
- and that $-\pi/2 \leq \theta \leq \pi/2.$
$\underline{\text{Preliminary Result-1}}$
$\sin(\pi - M) = \sin(\pi)\cos(M) - \sin(M)\cos(\pi) = \sin(M).$
The computation of $\theta$ may then be broken down into $(3)$ cases:
$\underline{\text{Case 1:} ~~~0 \leq M \leq \pi/2}$
Simply set $\theta = M.$
$\underline{\text{Case 2:} ~~~\pi/2 < M < 3\pi/2}$
Using Preliminary Result-1,
simply set $\theta = (\pi - M).$
$\underline{\text{Case 3:} ~~~3\pi/2 \leq M < 2\pi}$
Simply set $\theta = M - 2\pi.$
