91Ó°ÊÓ

First, let's define variables for the inverse sine functions. Let $$x = \sin ^{-1}(y)$$, then by definition, $$\sin(x) = y$$. Similarly, let $$z = \sin ^{-1}(-y)$$, then $$\sin(z) = -y$$. So the equation we want to prove becomes: $$x + z = 0$$

We know the sine addition formula is $$\sin(a + b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$ Let's apply this formula to our variables $$x$$ and $$z$$: $$\sin(x + z) = \sin(x) \cos(z) + \cos(x) \sin(z)$$

Now, we substitute the values from step 1 into the equation from step 2. $$\sin(x + z) = y\cos(z) + \cos(x)(-y)$$

Now we want to prove that $$x + z = 0$$ For that, we first find the value of $$\sin(x + z)$$ We already know $$\sin(x) = y$$ $$\sin(z) = -y$$ So, based on sin subtraction formula $$\sin(x - z) = \sin(x)\cos(z) - \cos(x)\sin(z)$$ Now if we sum sin(x + z) and sin(x - z), $$\sin(x + z) + \sin(x - z) = (y\cos(z) - \cos(x)(-y)) + (y\cos(z) + \cos(x)(-y))$$ Simplifying, we get $$\sin(x + z) + \sin(x - z) = 2y\cos(z)$$ We also know that $$\sin(x - z) = \sin(x + (-z)) = y\cos(-z) = y\cos(z)$$ Therefore, $$\sin(x + z) + \sin(x - z) = 2y\cos(z) = 2\sin(x - z)$$ Which implies, $$\sin(x + z) = \sin(x - z)$$ Now, since both sine values are equal, it implies that either: 1. $$x + z = x - z + 2k\pi$$, where k is an integer. 2. $$x + z = \pi - (x - z) + 2k\pi$$, where k is an integer. The first case simplifies to $$z = -z + 2k\pi$$, which can only happen if $$k$$ is an integer and $$z = k\pi$$, which is not possible since $$z = \sin ^{-1}(-y)$$, the range is $$-\frac{\pi}{2} \leq z \leq \frac{\pi}{2}$$ The second case simplifies to $$2z = (\pi - 2x) + 2k\pi$$, but as discussed above, $$z = \sin ^{-1}(-y)$$ and $$x = \sin ^{-1}(y)$$ so the equation becomes: $$2\sin ^{-1}(-y) = (\pi - 2\sin ^{-1}(y)) + 2k\pi$$ We see this equation is also not true for the given range of angles since the inverse sine function has a limited range. Thus, we're left with the scenario that $$x + z = 0$$ which proves the original identity: $$\sin ^{-1} y+\sin ^{-1}(-y)=0$$