evaluate the following expressions. your answer must be an angle -π/2 ≤ θ < π in radians, written as a…

evaluate the following expressions. your answer must be an angle -π/2 ≤ θ < π in radians, written as a multiple of π. note that π is already provided in the answer so you simply have to fill in the appropriate multiple. e.g. if the answer is π/2 you should enter 1/2. do not use decimal answers. write the answer as a fraction or integer. sin⁻¹(sin(3π/4)) = □ π sin⁻¹(sin(-π/3)) = □ π cos⁻¹(cos(π/6)) = □ π cos⁻¹(cos(7π/6)) = □ π

evaluate the following expressions. your answer must be an angle -π/2 ≤ θ < π in radians, written as a multiple of π. note that π is already provided in the answer so you simply have to fill in the appropriate multiple. e.g. if the answer is π/2 you should enter 1/2. do not use decimal answers. write the answer as a fraction or integer. sin⁻¹(sin(3π/4)) = □ π sin⁻¹(sin(-π/3)) = □ π cos⁻¹(cos(π/6)) = □ π cos⁻¹(cos(7π/6)) = □ π

Answer

Explanation:

Step1: Recall inverse - sine property

The property of the inverse - sine function (y = \sin^{-1}(x)) is that (\sin^{-1}(\sin\theta)=\theta) when (-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}). For (\sin^{-1}(\sin(\frac{3\pi}{4}))), since (\frac{3\pi}{4}\gt\frac{\pi}{2}), we use the identity (\sin\theta=\sin(\pi - \theta)). Here, (\sin(\frac{3\pi}{4})=\sin(\pi-\frac{3\pi}{4})=\sin(\frac{\pi}{4})), so (\sin^{-1}(\sin(\frac{3\pi}{4}))=\frac{\pi}{4}).

Step2: Recall inverse - sine property for negative angle

For (\sin^{-1}(\sin(-\frac{\pi}{3}))), since (-\frac{\pi}{2}\leq-\frac{\pi}{3}\leq\frac{\pi}{2}), then (\sin^{-1}(\sin(-\frac{\pi}{3})) =-\frac{\pi}{3}).

Step3: Recall inverse - cosine property

The property of the inverse - cosine function (y=\cos^{-1}(x)) is that (\cos^{-1}(\cos\theta)=\theta) when (0\leq\theta\leq\pi). For (\cos^{-1}(\cos(\frac{\pi}{6}))), since (0\leq\frac{\pi}{6}\leq\pi), then (\cos^{-1}(\cos(\frac{\pi}{6}))=\frac{\pi}{6}).

Step4: Recall inverse - cosine property for (\theta\gt\pi)

For (\cos^{-1}(\cos(\frac{7\pi}{6}))), since (\frac{7\pi}{6}\gt\pi), we use the identity (\cos\theta=\cos(2\pi - \theta)) or (\cos\theta=\cos(-\theta)). Also, (\cos\theta=\cos(2k\pi\pm\theta)), (k\in\mathbb{Z}). We know that (\cos(\frac{7\pi}{6})=\cos(2\pi-\frac{7\pi}{6})=\cos(\frac{5\pi}{6})), and since (0\leq\frac{5\pi}{6}\leq\pi), then (\cos^{-1}(\cos(\frac{7\pi}{6}))=\frac{5\pi}{6}).

Answer:

(\sin^{-1}(\sin(\frac{3\pi}{4}))=\frac{1}{4}\pi) (\sin^{-1}(\sin(-\frac{\pi}{3}))=-\frac{1}{3}\pi) (\cos^{-1}(\cos(\frac{\pi}{6}))=\frac{1}{6}\pi) (\cos^{-1}(\cos(\frac{7\pi}{6}))=\frac{5}{6}\pi)