Question

Prove the following identities. $$\sin ^{-1} y+\sin ^{-1}(-y)=0$$

Short answer

Question: Prove the identity \(\sin^{-1}y + \sin^{-1}(-y) = 0\). Short Answer: To prove the given identity, we can use the properties of the inverse sine function and the fact that the sine function is an odd function. Since we know that \(\sin(-x) = -\sin(x)\), we can use this property to show that \(-\sin^{-1}(\sin(x)) = \sin^{-1}(-y)\). This results in \(-\sin^{-1}(y) = \sin^{-1}(-y)\), which helps us confirm the identity, as \(\sin^{-1}(y) + \sin^{-1}(-y) = \sin^{-1}(y) - \sin^{-1}(y) = 0\).

Step-by-step solution

Understand inverse sine function

The inverse sine function, denoted as \(\sin^{-1}(x)\), is the function that "undoes" the sine function. In other words, if \(\sin(x) = y\), then \(\sin^{-1}(y) = x\). Note that the range of the inverse sine function is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

Recall the property of the sine function in relation to negative angles

Recall that the sine function has the property that \(\sin(-x) = -\sin(x)\). In other words, the sine function is an odd function. This means that the sine function is symmetric with respect to the origin, which has important implications for the inverse sine function.

Understand symmetry for the inverse sine function

Given that the sine function is odd and symmetric with respect to the origin, we know that for any angle \(x\) in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\), we have that \(\sin^{-1}(-\sin(x)) = -\sin^{-1}(\sin(x))\).

Apply the identity to the given expression

We want to prove that: $$\sin ^{-1} y+\sin ^{-1}(-y)=0$$ Let \(x = \sin^{-1}(y)\), so \(\sin(x) = y\). We know from the properties of the sine function that \(\sin(-x) = -\sin(x) = -y\). Now, we use the inverse sine function on both sides of the equation: $$-\sin^{-1}(\sin(x)) = \sin^{-1}(-y)$$ Since \(x = \sin^{-1}(y)\), we have: $$-\sin^{-1}(y) = \sin^{-1}(-y)$$

Combine the expressions

Finally, we can substitute this expression back into the original problem: $$\sin^{-1}(y) + \sin^{-1}(-y) = 0$$ Since we found that \(-\sin^{-1}(y) = \sin^{-1}(-y)\), the expression becomes: $$\sin^{-1}(y) - \sin^{-1}(y) = 0$$ Which simplifies to: $$0 = 0$$ This confirms the given identity: $$\sin ^{-1} y+\sin ^{-1}(-y)=0$$