Question

sin(arccos(−11/12))

Answer 1

To find \(\sin(\arccos(-11/12))\), we can use the identity that relates sine and cosine.

Given \( \theta = \arccos(-11/12) \), we have:

\[
\cos(\theta) = -\frac{11}{12}
\]

We can find \(\sin(\theta)\) using the Pythagorean identity:

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

Substituting \(\cos(\theta)\):

\[
\sin^2(\theta) + \left(-\frac{11}{12}\right)^2 = 1
\]

Calculating \(\left(-\frac{11}{12}\right)^2\):

\[
\sin^2(\theta) + \frac{121}{144} = 1
\]

Now, solve for \(\sin^2(\theta)\):

\[
\sin^2(\theta) = 1 - \frac{121}{144}
\]

Convert 1 to a fraction with the same denominator:

\[
1 = \frac{144}{144}
\]

Now the equation becomes:

\[
\sin^2(\theta) = \frac{144}{144} - \frac{121}{144} = \frac{23}{144}
\]

Taking the square root:

\[
\sin(\theta) = \pm \sqrt{\frac{23}{144}} = \pm \frac{\sqrt{23}}{12}
\]

Since \(\theta = \arccos(-11/12)\) is in the range \([0, \pi]\), where sine is non-negative, we take the positive value:

\[
\sin(\arccos(-11/12)) = \frac{\sqrt{23}}{12}
\]

Thus, the final answer is:

\[
\sin(\arccos(-11/12)) = \frac{\sqrt{23}}{12}
\]