Question

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

Short answer

Question: Prove the following trigonometric identity: \(\sin^{-1}y + \sin^{-1}(-y) = 0\). Solution: We can prove this identity by following these steps: Step 1: Define two angles and rewrite the identity in terms of sine functions. Let \(x_1 = \sin^{-1}y\) and \(x_2 = \sin^{-1}(-y)\). Then, $$\sin x_1 = y$$ $$\sin x_2 = -y$$ Step 2: Use the sine addition formula to find a relationship between the angles. $$\sin(x_1 + x_2) = \sin x_1 \cos x_2 + \cos x_1 \sin x_2$$ Substitute the expressions for \(\sin x_1\) and \(\sin x_2\): $$\sin(x_1 + x_2) = y \cos x_2 - \cos x_1 y$$ Factor out \(y\): $$\sin(x_1 + x_2) = y(\cos x_2 - \cos x_1)$$ Step 3: Show that the sum of the angles is a multiple of \(\pi\). $$\sin(x_1 + x_2) = 0$$ $$x_1 + x_2 = k\pi$$ where \(k\) is an integer. Step 4: Substitute back in the inverse sine functions. $$\sin^{-1}y + \sin^{-1}(-y) = k\pi$$ Since the range of the inverse sine function is \(-\frac{\pi}{2} \leq \sin^{-1}y \leq \frac{\pi}{2}\), the only integer value of \(k\) that satisfies the range is \(k = 0\). Therefore, $$\sin^{-1}y + \sin^{-1}(-y) = 0$$ This proves the given identity.

Step-by-step solution

Consider the inverse sine function

Recall the definition of the inverse sine function: If \(\sin x = y\), then \(\sin^{-1}y = x\). We will use this property to convert the inverse sine functions into regular sine functions by declaring two angles. Let \(x_1 = \sin^{-1}y\) and \(x_2 = \sin^{-1}(-y)\). Then, by the definition of the inverse sine function, we have: $$\sin x_1 = y$$ $$\sin x_2 = -y$$

Use the sine addition formula

Now, we want to find a relationship between \(x_1\) and \(x_2\). We can use the sine addition formula, which states that \(\sin(a + b) = \sin a \cos b + \cos a \sin b\). In our case, we want to find the sum of \(x_1\) and \(x_2\): $$\sin(x_1 + x_2) = \sin x_1 \cos x_2 + \cos x_1 \sin x_2$$ Substitute the expressions we found for \(\sin x_1\) and \(\sin x_2\): $$\sin(x_1 + x_2) = y \cos x_2 - \cos x_1 y$$ Factor out \(y\): $$\sin(x_1 + x_2) = y(\cos x_2 - \cos x_1)$$

Show that the sum of the angles is a multiple of \(\pi\)

We want to show that the sum of the angles results in the sine value being zero: $$\sin(x_1 + x_2) = 0$$ This means that \(x_1 + x_2\) must be a multiple of \(\pi\): $$x_1 + x_2 = k\pi$$ where \(k\) is an integer.

Substitute back in the inverse sine functions

Now we need to convert this back into our inverse sine functions. Recall that \(x_1 = \sin^{-1}y\) and \(x_2 = \sin^{-1}(-y)\): $$\sin^{-1}y + \sin^{-1}(-y) = k\pi$$ Since the inverse sine function's range is from \(-\frac{\pi}{2} \leq \sin^{-1}y \leq \frac{\pi}{2}\), the only integer value \(k\) that satisfies the range is \(k = 0\). Thus, $$\sin^{-1}y + \sin^{-1}(-y) = 0\pi = 0$$ We have successfully proven the given identity.