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Chapter 5: Problem 38

Verify each identity. (a) \(\arcsin (-x)=-\arcsin x, \quad|x| \leq 1\) (b) \(\arccos (-x)=\pi-\arccos x, \quad|x| \leq 1\)

Short Answer

Expert verified
For \(|x| \leq 1\), \(\arcsin(-x)=-\arcsin x\) and \(\arccos(-x) = \pi-\arccos x\) are both valid mathematical identities.

Step by step solution

01

Verifying Identity for \(\arcsin (-x)=-\arcsin x\)

Firstly, consider the left-hand side, \(\arcsin(-x)\). If \(\sin y = -x\), then \(-\sin y = x\). As sine is an odd function, meaning \(\sin(-y) = -\sin(y)\), so \(\sin(-y) = x\), meaning \(-y = \arcsin x\). Hence, \(\arcsin(-x) = -\arcsin x\) is validated for any \(x\) such that \(|x| \leq 1\).
02

Verifying Identity for \(\arccos(-x) = \pi-\arccos x\)

For the second identity, start with the right-hand side, \(\pi - \arccos x\). If \(\cos y = x\), then we have \(\cos(\pi - y) = -x\). As cos is an even function, meaning \(\cos(-y) = \cos(y)\), and here \(y = \pi - \arccos x\), which leads us \(\cos(-y) = -x \), this means \(-y = \arccos(-x)\). Hence, \(\pi - \arccos x = \arccos(-x)\) is validated for any \(x\) such that \(|x| \leq 1\).

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