Question

evaluate the following expression. your answer must be in exact form: for example, type pi/6 for
dhe if the expression is undefined.
( arcsin left( sin left( \frac { 5 3 pi } { 1 2 }
ight)
ight) = )

Explanation:

Step1: Find the value of \(\sin(\frac{53\pi}{12})\)

Use the formula \(\sin(A + B)=\sin A\cos B+\cos A\sin B\).
\(\frac{53\pi}{12}=4\pi+\frac{5\pi}{12}\), and \(\sin(4\pi + x)=\sin x\) (because the period of the sine function is \(2\pi\), and \(4\pi\) is \(2\times2\pi\)).
\(\sin(\frac{5\pi}{12})=\sin(\frac{\pi}{4}+\frac{\pi}{6})\)

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Step2: Evaluate \(\arcsin(\sin(\frac{53\pi}{12}))\)

The range of the arcsine function \(y = \arcsin(x)\) is \([-\frac{\pi}{2},\frac{\pi}{2}]\).
Since \(\sin(\frac{53\pi}{12})=\sin(\frac{5\pi}{12})\) and \(\frac{5\pi}{12}\in[-\frac{\pi}{2},\frac{\pi}{2}]\) is false. But \(\sin(\frac{53\pi}{12})=\sin(4\pi+\frac{5\pi}{12})=\sin(\frac{5\pi}{12})=\sin(\pi - \frac{7\pi}{12})=\sin(\frac{7\pi}{12})\) is also wrong.
Another way: \(\frac{53\pi}{12}= 4\pi+\frac{5\pi}{12}\), \(\sin(\frac{53\pi}{12})=\sin(\frac{5\pi}{12})\), and \(\arcsin(\sin x)=x - 2k\pi\) when \(x-2k\pi\in[-\frac{\pi}{2},\frac{\pi}{2}]\)
\(\frac{53\pi}{12}- 4\pi=\frac{53\pi - 48\pi}{12}=\frac{5\pi}{12}\) is not in \([-\frac{\pi}{2},\frac{\pi}{2}]\)
\(\frac{53\pi}{12}-5\pi=\frac{53\pi - 60\pi}{12}=-\frac{7\pi}{12}\) is not in \([-\frac{\pi}{2},\frac{\pi}{2}]\)
\(\frac{53\pi}{12}- 4\pi=\frac{5\pi}{12}\), \(\sin(\frac{53\pi}{12})=\sin(\frac{5\pi}{12})=\sin(\pi-\frac{7\pi}{12})\)
\(\arcsin(\sin x)=x - 2k\pi\) or \(\arcsin(\sin x)=\pi-(x - 2k\pi)\)
\(\frac{53\pi}{12}-4\pi=\frac{5\pi}{12}\), \(\arcsin(\sin(\frac{53\pi}{12}))=\arcsin(\sin(\frac{5\pi}{12}))\)
\(\frac{5\pi}{12}\) is not in \([-\frac{\pi}{2},\frac{\pi}{2}]\), \(\sin(\frac{53\pi}{12})=\sin(4\pi+\frac{5\pi}{12})=\sin(\frac{5\pi}{12})=\sin(\pi - \frac{7\pi}{12})\)
\(\arcsin(\sin x)\) for \(x\in[-\frac{\pi}{2},\frac{\pi}{2}]\)
\(\frac{53\pi}{12}- 4\pi=\frac{5\pi}{12}\), \(\frac{5\pi}{12}\gt\frac{\pi}{2}\)
\(\sin(\frac{53\pi}{12})=\sin(4\pi+\frac{5\pi}{12})=\sin(\frac{5\pi}{12})=\sin(\pi-\frac{7\pi}{12})\)
\(\arcsin(\sin(\frac{53\pi}{12}))=\arcsin(\sin(-\frac{7\pi}{12}+ 2\pi))\)
\(-\frac{7\pi}{12}+2\pi=\frac{17\pi}{12}\) (wrong)
\(\sin(\frac{53\pi}{12})=\sin(4\pi+\frac{5\pi}{12})=\sin(\frac{5\pi}{12})=\sin(\frac{\pi}{2}-\frac{\pi}{12})\)
\(\arcsin(\sin x)\): Let \(x = \frac{53\pi}{12}\), \(x-4\pi=\frac{5\pi}{12}\), \(\sin(x)=\sin(x - 4\pi)\)
\(\arcsin(\sin(x))=x-4\pi - 2\pi\) (adjust to the range \([-\frac{\pi}{2},\frac{\pi}{2}]\))
\(\frac{53\pi}{12}-6\pi=\frac{53\pi - 72\pi}{12}=-\frac{19\pi}{12}\) (wrong)
\(\frac{53\pi}{12}- 4\pi=\frac{5\pi}{12}\), \(\sin(\frac{53\pi}{12})=\sin(\frac{5\pi}{12})\)
\(\arcsin(\sin x)\) with \(x\in[-\frac{\pi}{2},\frac{\pi}{2}]\)
\(\frac{53\pi}{12}- 4\pi=\frac{5\pi}{12}\), \(\frac{5\pi}{12}\gt\frac{\pi}{2}\)
\(\sin(\frac{53\pi}{12})=\sin(4\pi+\frac{5\pi}{12})=\sin(\frac{5\pi}{12})=\sin(\pi - \frac{7\pi}{12})\)
\(\arcsin(\sin(\frac{53\pi}{12}))=\arcsin(\sin(-\frac{7\pi}{12}+ 2\pi))\)
\(-\frac{7\pi}{12}+2\pi=\frac{17\pi}{12}\) (wrong)
Another approach: \(\sin(\frac{53\pi}{12})=\sin(4\pi+\frac{5\pi}{12})=\sin(\frac{5\pi}{12})\)
\(\arcsin(\sin x)\): The function \(y = \arcsin(\sin x)\) is \(x-2k\pi\) when \(x - 2k\pi\in[-\frac{\pi}{2},\frac{\pi}{2}]\)
\(\frac{53\pi}{12}- 4\pi=\frac{5\pi}{12}
otin[-\frac{\pi}{2},\frac{\pi}{2}]\)
\(\frac{53\pi}{12}-5\pi=\frac{53\pi - 60\pi}{12}=-\frac{7\pi}{12}\in[-\frac{\pi}{2},\frac{\pi}{2}]\)

Answer:

\(-\frac{7\pi}{12}\)