problems 7 - 12, true or false. do not use a calculator. give a reason for your answer.\n7…

Explanation:

Step1: Recall the property of inverse - sine function

The inverse - sine function (y = \sin^{-1}(x)) has a domain ([- 1,1]) and range (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]). For (y=\sin^{-1}(\sin(x))), if (x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]), then (\sin^{-1}(\sin(x)) = x). Since (-\frac{\pi}{10}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]), (\sin^{-1}\left[\sin\left(-\frac{\pi}{10}\right)\right]=-\frac{\pi}{10}). So, the statement in problem 7 is True.

Step2: Recall the property of inverse - cosine function

The inverse - cosine function (y = \cos^{-1}(x)) has a domain ([-1,1]) and range ([0,\pi]). For (y = \cos^{-1}(\cos(x))), if (x\in[0,\pi]), then (\cos^{-1}(\cos(x))=x). But for (x = 2), since (2\notin[-1,1]), the expression (\cos^{-1}(2)) is not defined. So, the statement in problem 8 is False.

Step3: Recall the property of inverse - cosine function

First, (\cos(-\frac{\pi}{4})=\cos(\frac{\pi}{4})) because (\cos(-x)=\cos(x)). The inverse - cosine function (y=\cos^{-1}(x)) has range ([0,\pi]), and (\cos^{-1}\left[\cos\left(-\frac{\pi}{4}\right)\right]=\cos^{-1}\left[\cos\left(\frac{\pi}{4}\right)\right]=\frac{\pi}{4}\neq-\frac{\pi}{4}). So, the statement in problem 9 is False.

Step4: Recall the property of inverse - tangent function

The inverse - tangent function (y = \tan^{-1}(x)) has a domain ((-\infty,\infty)) and range (\left(-\frac{\pi}{2},\frac{\pi}{2}\right)). Since (-\frac{\pi}{3}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)), (\tan^{-1}\left[\tan\left(-\frac{\pi}{3}\right)\right]=-\frac{\pi}{3}). So, the statement in problem 10 is True.

Step5: Recall the property of inverse - sine function

The inverse - sine function (y=\sin^{-1}(x)) has range (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]). We know that (\sin\left(\frac{2\pi}{3}\right)=\sin\left(\pi - \frac{\pi}{3}\right)=\sin\left(\frac{\pi}{3}\right)). And (\sin^{-1}\left[\sin\left(\frac{2\pi}{3}\right)\right]=\sin^{-1}\left[\sin\left(\frac{\pi}{3}\right)\right]=\frac{\pi}{3}\neq\frac{2\pi}{3}) because (\frac{2\pi}{3}\notin\left[-\frac{\pi}{2},\frac{\pi}{2}\right]). So, the statement in problem 11 is False.

Step6: Recall the property of inverse - tangent function

The inverse - tangent function (y = \tan^{-1}(x)) has a domain ((-\infty,\infty)) and range (\left(-\frac{\pi}{2},\frac{\pi}{2}\right)). For any (x\in(-\infty,\infty)), (\tan\left[\tan^{-1}(x)\right]=x). Since (-\frac{1}{2}\in(-\infty,\infty)), (\tan\left[\tan^{-1}\left(-\frac{1}{2}\right)\right]=-\frac{1}{2}). So, the statement in problem 12 is True.

Answer:

7. True
8. False
9. False
10. True
11. False
12. True