Cos Arcsin Value 6729Ba

1. **State the problem:** Find the value of $$\cos\left(\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right)$$. 2. **Recall the formula and important rules:** If $$y = \sin^{-1}(x)$$, then $$\sin(y) = x$$ and $$y$$ lies in $$[-\frac{\pi}{2}, \frac{\pi}{2}]$$ (the principal range of arcsine). We want to find $$\cos(y)$$ where $$\sin(y) = -\frac{\sqrt{2}}{2}$$. Use the Pythagorean identity: $$\cos^2(y) + \sin^2(y) = 1$$ So, $$\cos(y) = \pm \sqrt{1 - \sin^2(y)}$$. 3. **Calculate $$\cos(y)$$:** $$\sin(y) = -\frac{\sqrt{2}}{2}$$ Square it: $$\sin^2(y) = \left(-\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2}$$ Then, $$\cos(y) = \pm \sqrt{1 - \frac{1}{2}} = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2}$$ 4. **Determine the sign of $$\cos(y)$$:** Since $$y = \sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)$$, $$y$$ is in the interval $$[-\frac{\pi}{2}, \frac{\pi}{2}]$$ (quadrants IV and I). Given $$\sin(y) < 0$$, $$y$$ must be in quadrant IV where cosine is positive. Therefore, $$\cos(y) = +\frac{\sqrt{2}}{2}$$ 5. **Final answer:** $$\cos\left(\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right) = \frac{\sqrt{2}}{2}$$