Question

rewrite each of the following equations using the corresponding trigonometric function. then determine the exact value of the inverse trigonometric fun(a) ( t=arcsin left(\frac{sqrt{2}}{2}
ight) )this means thatwhere ( t ) lies in the intervalthus ( arcsin left(\frac{sqrt{2}}{2}
ight)=$(b) ( t=arcsin left(-\frac{sqrt{2}}{2}
ight) )this means thatwhere ( t ) lies in the intervalthus ( arcsin left(-\frac{sqrt{2}}{2}
ight)=$question help: message instructorsubmit question

Explanation:

Step1: Recall the definition of arcsin

The equation \(t = \arcsin(x)\) means \(\sin(t)=x\), where \(t\in[-\frac{\pi}{2},\frac{\pi}{2}]\)

Part (a)

Step2: Apply the definition to \(t = \arcsin(\frac{\sqrt{2}}{2})\)

Since \(t = \arcsin(\frac{\sqrt{2}}{2})\), then \(\sin(t)=\frac{\sqrt{2}}{2}\), and \(t\in[-\frac{\pi}{2},\frac{\pi}{2}]\).
We know that \(\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\) and \(\frac{\pi}{4}\in[-\frac{\pi}{2},\frac{\pi}{2}]\)

Part (b)

Step3: Apply the definition to \(t = \arcsin(-\frac{\sqrt{2}}{2})\)

Since \(t = \arcsin(-\frac{\sqrt{2}}{2})\), then \(\sin(t)=-\frac{\sqrt{2}}{2}\), and \(t\in[-\frac{\pi}{2},\frac{\pi}{2}]\)
We know that \(\sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}\) and \(-\frac{\pi}{4}\in[-\frac{\pi}{2},\frac{\pi}{2}]\)

Answer:

• (a) This means that \(\sin(t)=\frac{\sqrt{2}}{2}\), where \(t\) lies in the interval \([-\frac{\pi}{2},\frac{\pi}{2}]\). Thus \(\arcsin(\frac{\sqrt{2}}{2})=\frac{\pi}{4}\)
• (b) This means that \(\sin(t)=-\frac{\sqrt{2}}{2}\), where \(t\) lies in the interval \([-\frac{\pi}{2},\frac{\pi}{2}]\). Thus \(\arcsin(-\frac{\sqrt{2}}{2})=-\frac{\pi}{4}\)