Question

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

Short answer

The two identities \(\arcsin(-x) = -\arcsin x\) and \(\arccos(-x) = \pi - \arccos x\) are correct for all \(|x| \leq 1\).

Step-by-step solution

Identity for arcsin simplification

Let's denote \(y = \arcsin(-x)\). Therefore, by definition of the inverse sine, we have \(\sin(y) = -x\), for \(|y| \leq \frac{\pi}{2}\). Then, applying the definition of sin for the negative angle we can rewrite \(\sin(y) = \sin(-(-y)) = -\sin(-y)\). So, \(-x = -\sin(-y)\), which gives \(x = \sin(-y)\). As a result, \(-y = \arcsin(x)\) because \(-y\) is the angle whose sine is \(x\). Since \(\arcsin(x)\) is bound to the range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\) and \(-y\) falls in the same range, we can conclude that \(-y = \arcsin(x)\). Thus, the identity \(\arcsin (-x)=-\arcsin x\) is verified.

Identity for arccos simplification

Now, let's consider \(y = \arccos(-x)\). Applying the properties of cosine function, we get \(\cos(y) = -x\), for \(0 \leq y \leq \pi\). According to the properties of cosine function, \(\cos(y) = \cos(\pi -y)\). So \(-x = \cos(\pi - y)\), which leads to \(x = - \cos(\pi -y)\) or \(x= \cos(y-\pi)\). Since \(\arccos(x)\) is in the range \(0 \leq y \leq \pi\), the only possible value for \(y\) is \(\pi - \arccos(x)\). Thus, the identity \(\arccos (-x)=\pi-\arccos x\) is verified.