Questions: Find the exact value of cos(sin^(-1)(-7/8)).

Solution Steps

To find the exact value of \(\cos \left(\sin ^{-1}\left(-\frac{7}{8}\right)\right)\), we can use the relationship between the trigonometric functions and their inverses. Specifically, if \(\theta = \sin^{-1}\left(-\frac{7}{8}\right)\), then \(\sin(\theta) = -\frac{7}{8}\). We can use the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\) to find \(\cos(\theta)\).

1. Let \(\theta = \sin^{-1}\left(-\frac{7}{8}\right)\), so \(\sin(\theta) = -\frac{7}{8}\).
2. Use the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\) to solve for \(\cos(\theta)\).
3. Since \(\theta\) is in the range of \(\sin^{-1}\), which is \([- \frac{\pi}{2}, \frac{\pi}{2}]\), \(\cos(\theta)\) will be positive.

Let \( \theta = \sin^{-1}\left(-\frac{7}{8}\right) \). This implies that \( \sin(\theta) = -\frac{7}{8} \).

Using the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), we can find \( \cos(\theta) \): \[ \sin^2(\theta) = \left(-\frac{7}{8}\right)^2 = \frac{49}{64} \] Substituting into the identity: \[ \frac{49}{64} + \cos^2(\theta) = 1 \]

Rearranging gives: \[ \cos^2(\theta) = 1 - \frac{49}{64} = \frac{64}{64} - \frac{49}{64} = \frac{15}{64} \] Taking the square root: \[ \cos(\theta) = \sqrt{\frac{15}{64}} = \frac{\sqrt{15}}{8} \] Since \( \theta \) is in the range of \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), \( \cos(\theta) \) is positive.

Final Answer