Question

Evaluate cosine squared left parenthesis Start Fraction pi over 12 End Fraction right parenthesis.
(1 point)
Responses

one fourth
Image with alt text: one fourth

Start Fraction 2 plus Start Root 3 End Root over 4 End Fraction
Image with alt text: Start Fraction 2 plus Start Root 3 End Root over 4 End Fraction

3 halves
Image with alt text: 3 halves

Start Fraction 2 minus Start Root 3 End Root over 4 End Fraction

Answer 1

To evaluate \(\cos^2\left(\frac{\pi}{12}\right)\), we can first find \(\cos\left(\frac{\pi}{12}\right)\).

We know that \(\frac{\pi}{12} = 15^\circ\). We can use the cosine angle subtraction formula:

\[
\cos\left(\frac{\pi}{12}\right) = \cos\left(15^\circ\right) = \cos(45^\circ - 30^\circ)
\]

Using the cosine subtraction formula:

\[
\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)
\]

So,

\[
\cos\left(15^\circ\right) = \cos(45^\circ)\cos(30^\circ) + \sin(45^\circ)\sin(30^\circ)
\]
\[
= \left(\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\cdot\frac{1}{2}\right)
\]
\[
= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}
\]
\[
= \frac{\sqrt{6} + \sqrt{2}}{4}
\]

Now we need \(\cos^2\left(15^\circ\right)\):

\[
\cos^2\left(\frac{\pi}{12}\right) = \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2
\]
\[
= \frac{(\sqrt{6} + \sqrt{2})^2}{16}
\]
\[
= \frac{6 + 2 + 2\sqrt{12}}{16}
\]
\[
= \frac{8 + 4\sqrt{3}}{16}
\]
\[
= \frac{2 + \sqrt{3}}{4}
\]

Thus,

\[
\cos^2\left(\frac{\pi}{12}\right) = \frac{2 + \sqrt{3}}{4}
\]

So, the correct answer is:

\(\frac{2 + \sqrt{3}}{4}\)