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Chapter 19: Problem 1889

If \(\sin ^{-1} x-\cos ^{-1} x<0\) then (a) \(-1 \leq x<(1 / \sqrt{2})\) (b) \(-1

Short Answer

Expert verified
The correct answer is (b) \(-1 < x < 0\).

Step by step solution

01

Rewrite the inequality using the property of inverse trigonometric functions

We know that \(\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}\). Therefore, we can rewrite the given inequality as: \[\sin ^{-1} x - \frac{\pi}{2} < -\cos ^{-1} x\]
02

Apply sine function to both sides of the inequality

Take the sine of both sides of the inequality: \[\sin\left(\sin ^{-1} x - \frac{\pi}{2}\right) < \sin(-\cos ^{-1} x)\] Since we know that \(\sin(\alpha - \frac{\pi}{2}) = \cos(\alpha)\) for any angle \(\alpha\), we can rewrite the left-hand side of the inequality as: \[\cos(\sin ^{-1} x) < \sin(-\cos ^{-1} x)\]
03

Use the property of sin(-x) = -sin(x) for cosine inequality

Since the sine function is an odd function, we have \(\sin(-x) = -\sin(x)\). Therefore, we can rewrite the right-hand side of the inequality as: \[\cos(\sin ^{-1} x) < -\sin(\cos ^{-1} x)\]
04

Apply Pythagorean identity to simplify the inequality

Using the Pythagorean identity \(\sin^2(x) + \cos^2(x) = 1\), we can express \(\sin(\cos^{-1}(x))\) and \(\cos(\sin^{-1}(x))\) in terms of x. We have: \[\sin(\cos^{-1}(x)) = \sqrt{1 - x^2}\] and \[\cos(\sin^{-1}(x)) = \sqrt{1 - x^2}\] Substitute these expressions back into the inequality: \[\sqrt{1 - x^2} < -\sqrt{1 - x^2}\]
05

Solve the inequality for x

Since both sides of the inequality involve a square root, we can square both sides to get rid of the square roots: \[(1 - x^2) < (1 - x^2)^2\] Which simplifies to: \[1 - x^2 < 1 - 2x^2+x^4\] Rearrange the inequality to have all terms on the left-hand side: \[x^4 - x^2 < 0\] Factor the left-hand side: \[x^2(x^2 - 1) < 0\] This inequality holds when x is in the interval: \[-1 < x < 0\] So, the correct answer is (b) \(-1 < x < 0\).

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