Question

csc\left(\frac{41\pi}{2}\
ight)
\bigcirc\\ undefined
\bigcirc\\ 0
\bigcirc\\ -1
\bigcirc\\ 1

Explanation:

Answer:

First, we find the coterminal angle of \(\frac{41\pi}{2}\) by subtracting multiples of \(2\pi\) (which is \(\frac{4\pi}{2}\)).

\(\frac{41\pi}{2}-20\times\frac{4\pi}{2}=\frac{41\pi}{2}-\frac{80\pi}{2}=\frac{41\pi - 80\pi}{2}=\frac{- 39\pi}{2}\) (this is not helpful, let's do it the other way, divide \(41\) by \(4\) since the period of \(\csc x\) (and \(\sin x\)) is \(2\pi=\frac{4\pi}{2}\), so the period in terms of \(\frac{\pi}{2}\) is \(4\)).

\(41\div4 = 10\) with a remainder of \(1\). So \(\frac{41\pi}{2}=10\times2\pi+\frac{\pi}{2}\) (wait, no, \(2\pi=\frac{4\pi}{2}\), so \(10\times\frac{4\pi}{2}=\frac{40\pi}{2}\), then \(\frac{41\pi}{2}=\frac{40\pi}{2}+\frac{\pi}{2}=20\pi+\frac{\pi}{2}\)).

We know that \(\csc x=\frac{1}{\sin x}\), and \(\sin(20\pi+\frac{\pi}{2})=\sin(\frac{\pi}{2}) = 1\) (because \(\sin(x + 2k\pi)=\sin x\) for integer \(k\)).

Then \(\csc(\frac{41\pi}{2})=\frac{1}{\sin(\frac{41\pi}{2})}=\frac{1}{\sin(20\pi+\frac{\pi}{2})}=\frac{1}{\sin(\frac{\pi}{2})}=\frac{1}{1} = 1\).

So the answer is \(1\) (the last option).

But following the output rules, we just give the final answer:

\(1\)