Science:Math Exam Resources/Courses/MATH100/December 2014/Question 03 (f)/Solution 1

The function $\arcsin$ has domain $[-1,1]$ and range $[-\pi /2,\pi /2]$, so our answer must lie in this range.

• We know that $\sin(x)=\sin(x-2\pi )$; so $\sin(9\pi /11)=\sin(31\pi /11)$.
• Also we know that $\sin(x)=\sin(\pi -x)$ and so, $\sin(9\pi /11)=\sin(2\pi /11)$.

Since $-\pi /2\leq 2\pi /11\leq \pi /2$, we conclude that

${\begin{aligned}\arcsin \left(\sin({\frac {31\pi }{11}})\right)&=\arcsin \left(\sin({\frac {2\pi }{11}})\right)\\&={\frac {2\pi }{11}}\end{aligned}}$