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Chapter 19: Problem 1870

\(\sin ^{-1}(\sin 2)+\sin ^{-1}(\sin 4)+\sin ^{-1}(\sin 6)=\) (a) \(\pi-12\) (b) 0 (c) 12 (d) \(12-\pi\)

Short Answer

Expert verified
(c) 12

Step by step solution

01

Find the values of the sine functions

We'll apply the sine function to the given arguments: \[\sin(2), \sin(4), \text{ and } \sin(6)\]
02

Apply inverse sine function

Now, we'll apply the inverse sine function to the results found in Step 1: \[\sin ^{-1}(\sin 2), \sin ^{-1}(\sin 4), \text{ and } \sin ^{-1}(\sin 6)\]
03

Use the properties of inverse sine

We know that for any angle \(x\) in the interval \([-90^\circ, 90^\circ]\) or \([\frac{-\pi}{2}, \frac{\pi}{2}]\), we can say: \[\sin ^{-1}(\sin x) = x\] In our case, angles 2, 4, and 6 are within the interval \([\frac{-\pi}{2}, \frac{\pi}{2}]\). Thus, we can apply the above property to each term, \[\sin ^{-1}(\sin 2) = 2\] \[\sin ^{-1}(\sin 4) = 4\] \[\sin ^{-1}(\sin 6) = 6\]
04

Sum up the results

Now sum up the results of the previous step: \[2 + 4 + 6 = 12\] So the expression \(\sin ^{-1}(\sin 2)+\sin ^{-1}(\sin 4)+\sin ^{-1}(\sin 6)\) equals 12. Answer: (c) 12

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