ections: evaluate the following expressions without a calculator. 1. $cos^{-1}(\frac{1}{2})$ 2…

Explanation:

Step1: Recall inverse - cosine range

The range of (y = \cos^{-1}(x)) is ([0,\pi]). We know that (\cos(\frac{\pi}{3})=\frac{1}{2}), so (\cos^{-1}(\frac{1}{2})=\frac{\pi}{3}).

Step2: Recall inverse - sine range

The range of (y=\sin^{-1}(x)) is ([-\frac{\pi}{2},\frac{\pi}{2}]). Since (\sin(-\frac{\pi}{4}) =-\frac{\sqrt{2}}{2}), then (\sin^{-1}(-\frac{\sqrt{2}}{2})=-\frac{\pi}{4}).

Step3: Recall inverse - tangent range

The range of (y = \tan^{-1}(x)) is ((-\frac{\pi}{2},\frac{\pi}{2})). As (\tan(\frac{\pi}{4}) = 1), so (\arctan(1)=\frac{\pi}{4}).

Step4: For (\arccos(-\frac{\sqrt{3}}{2}))

Since the range of (y=\cos^{-1}(x)) is ([0,\pi]) and (\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}), then (\arccos(-\frac{\sqrt{3}}{2})=\frac{5\pi}{6}).

Step5: For (\arcsin(-\frac{1}{2}))

With the range of (y = \sin^{-1}(x)) being ([-\frac{\pi}{2},\frac{\pi}{2}]) and (\sin(-\frac{\pi}{6})=-\frac{1}{2}), we have (\arcsin(-\frac{1}{2})=-\frac{\pi}{6}).

Step6: For (\tan^{-1}(-\frac{\sqrt{3}}{3}))

Given the range of (y=\tan^{-1}(x)) is ((-\frac{\pi}{2},\frac{\pi}{2})) and (\tan(-\frac{\pi}{6})=-\frac{\sqrt{3}}{3}), so (\tan^{-1}(-\frac{\sqrt{3}}{3})=-\frac{\pi}{6}).

Step7: For (\tan^{-1}(-1))

Since the range of (y = \tan^{-1}(x)) is ((-\frac{\pi}{2},\frac{\pi}{2})) and (\tan(-\frac{\pi}{4})=-1), then (\tan^{-1}(-1)=-\frac{\pi}{4}).

Step8: For (\cos^{-1}(\frac{\sqrt{3}}{2}))

As the range of (y=\cos^{-1}(x)) is ([0,\pi]) and (\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}), we get (\cos^{-1}(\frac{\sqrt{3}}{2})=\frac{\pi}{6}).

Step9: For (\arcsin(0))

With the range of (y = \sin^{-1}(x)) being ([-\frac{\pi}{2},\frac{\pi}{2}]) and (\sin(0) = 0), so (\arcsin(0)=0).

Step10: For (\cos^{-1}(\frac{\sqrt{2}}{2}))

Since the range of (y=\cos^{-1}(x)) is ([0,\pi]) and (\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}), then (\cos^{-1}(\frac{\sqrt{2}}{2})=\frac{\pi}{4}).

Step11: For (\arccos(-\frac{\sqrt{3}}{2})) (again)

The range of (y=\cos^{-1}(x)) is ([0,\pi]) and (\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}), so (\arccos(-\frac{\sqrt{3}}{2})=\frac{5\pi}{6}).

Step12: For (\arctan(\sqrt{3}))

Given the range of (y=\tan^{-1}(x)) is ((-\frac{\pi}{2},\frac{\pi}{2})) and (\tan(\frac{\pi}{3})=\sqrt{3}), so (\arctan(\sqrt{3})=\frac{\pi}{3}).

Answer:

1. (\frac{\pi}{3})
2. (-\frac{\pi}{4})
3. (\frac{\pi}{4})
4. (\frac{5\pi}{6})
5. (-\frac{\pi}{6})
6. (-\frac{\pi}{6})
7. (-\frac{\pi}{4})
8. (\frac{\pi}{6})
9. (0)
10. (\frac{\pi}{4})
11. (\frac{5\pi}{6})
12. (\frac{\pi}{3})