hw: practice for test 2 : chapter 6 due before 2pm\nscore: 10/19 answered: 10/19\nquestion 11\nevaluate the…

hw: practice for test 2 : chapter 6 due before 2pm\nscore: 10/19 answered: 10/19\nquestion 11\nevaluate the following expressions. your answer must be an exact angle in radians and in the interval 0,π. example: enter pi/6 for π/6.\n(a) cos⁻¹(−1)=\n(b) cos⁻¹(−√2/2)=\n(c) cos⁻¹(1)=\nquestion help: video 1 video 2 message instructor\nsubmit question

hw: practice for test 2 : chapter 6 due before 2pm\nscore: 10/19 answered: 10/19\nquestion 11\nevaluate the following expressions. your answer must be an exact angle in radians and in the interval 0,π. example: enter pi/6 for π/6.\n(a) cos⁻¹(−1)=\n(b) cos⁻¹(−√2/2)=\n(c) cos⁻¹(1)=\nquestion help: video 1 video 2 message instructor\nsubmit question

Answer

Explanation:

Step1: Recall inverse - cosine definition

The inverse - cosine function (y = \cos^{-1}(x)) gives the angle (\theta) in the interval ([0,\pi]) such that (\cos(\theta)=x).

Step2: Solve for (\cos^{-1}(-1))

We know that (\cos(\pi)= - 1) and (\pi\in[0,\pi]), so (\cos^{-1}(-1)=\pi).

Step3: Solve for (\cos^{-1}\left(-\frac{\sqrt{2}}{2}\right))

We know that (\cos\left(\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}) and (\frac{3\pi}{4}\in[0,\pi]), so (\cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)=\frac{3\pi}{4}).

Step4: Solve for (\cos^{-1}(1))

We know that (\cos(0) = 1) and (0\in[0,\pi]), so (\cos^{-1}(1)=0).

Answer:

(a) (\pi) (b) (\frac{3\pi}{4}) (c) (0)