91Ó°ÊÓ

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.