Question

cos(arcsin(−5/6))

Answer 1

To find \( \cos(\arcsin(-\frac{5}{6})) \), we can use the Pythagorean identity.

Let \( \theta = \arcsin\left(-\frac{5}{6}\right) \). This means that \( \sin(\theta) = -\frac{5}{6} \).

Using the Pythagorean identity, we have:

\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]

Substituting for \( \sin(\theta) \):

\[
\left(-\frac{5}{6}\right)^2 + \cos^2(\theta) = 1
\]

Calculating \( \left(-\frac{5}{6}\right)^2 \):

\[
\frac{25}{36} + \cos^2(\theta) = 1
\]

Now, solving for \( \cos^2(\theta) \):

\[
\cos^2(\theta) = 1 - \frac{25}{36}
\]

Converting 1 to a fraction with a denominator of 36:

\[
1 = \frac{36}{36}
\]

So we have:

\[
\cos^2(\theta) = \frac{36}{36} - \frac{25}{36} = \frac{11}{36}
\]

Taking the square root to find \( \cos(\theta) \):

\[
\cos(\theta) = \pm \sqrt{\frac{11}{36}} = \pm \frac{\sqrt{11}}{6}
\]

Next, we determine the correct sign for \( \cos(\theta) \). Since \( \theta = \arcsin\left(-\frac{5}{6}\right) \) is in the fourth quadrant (where sine is negative and cosine is positive), we take the positive root:

\[
\cos(\theta) = \frac{\sqrt{11}}{6}
\]

Thus, \( \cos(\arcsin(-\frac{5}{6})) = \frac{\sqrt{11}}{6} \).