Question

Use the fact that

$\cos \frac{\pi }{12}=\frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)$

to find$\sin \frac{\mathrm{\pi }}{24},\cos \frac{\mathrm{\pi }}{24}$.

Short answer

The value of $\sin \frac{\mathrm{\pi }}{24}$is $\frac{\sqrt{2}}{4}\sqrt{4-\sqrt{6}+\sqrt{2}}$.

The value of $\cos \frac{\mathrm{\pi }}{24}$is $\frac{\sqrt{2}}{4}\sqrt{4+\sqrt{6}+\sqrt{2}}$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\cos \frac{\pi }{12}=\frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)$

Assume $\sin \frac{\mathrm{\pi }}{24}$to be $\sin \theta$. Then,

$$\begin{gathered} \sin \frac{\mathrm{\pi }}{24}=\sin \theta \\ \theta =\frac{\mathrm{\pi }}{24} \end{gathered}$$

Also,$$\begin{gathered} \cos \frac{\pi }{12}=\cos \left(2\times \frac{\pi }{24}\right) \\ =\cos \left(2\theta \right) \end{gathered}$$

Step 2. Use the half angle formula.

Using the half angle formula,

${\sin }^2\theta =\frac{1-\cos \left(2\theta \right)}{2}$

Substitute the value of $\cos \left(2\theta \right)$in the equation,

role="math" localid="1646420828601" $$\begin{gathered} {\sin }^2\frac{\pi }{24}=\frac{1-\cos \left({\displaystyle \frac{\pi }{12}}\right)}{2} \\ \sin \frac{\pi }{24}=\sqrt{\frac{1-\cos \left({\displaystyle \frac{\pi }{12}}\right)}{2}} \\ \sin \frac{\pi }{24}=\sqrt{\frac{1-\left({\displaystyle \frac{\sqrt{6}+\sqrt{2}}{4}}\right)}{2}} \end{gathered}$$

Step 3. Simplify the expression on the right side.

Continuing the above equation,

$$\begin{gathered} \sin \frac{\pi }{24}=\sqrt{\frac{4-\left({\displaystyle \frac{\sqrt{6}+\sqrt{2}}{4}}\right)}{1}} \\ \sin \frac{\pi }{24}=\sqrt{\frac{4-\left({\displaystyle \sqrt{6}+\sqrt{2}}\right)}{8}} \\ \sin \frac{\pi }{24}=\frac{1}{2\sqrt{2}}\sqrt{4-\sqrt{6}+\sqrt{2}} \\ \sin \frac{\pi }{24}=\frac{\sqrt{2}}{4}\sqrt{4-\sqrt{6}+\sqrt{2}} \end{gathered}$$

Step 4. Use the half angle formula.

Using the half angle formula,

${\cos }^2\theta =\frac{1+\cos \left(2\theta \right)}{2}$

Substitute the value of $\cos \left(2\theta \right)$in the equation,

$$\begin{gathered} {\cos }^2\frac{\pi }{24}=\frac{1+\cos \left({\displaystyle \frac{\pi }{12}}\right)}{2} \\ \cos \frac{\pi }{24}=\sqrt{\frac{1+\left({\displaystyle \frac{\sqrt{6}+\sqrt{2}}{4}}\right)}{2}} \\ \cos \frac{\pi }{24}=\sqrt{\frac{4+\left({\displaystyle \frac{\sqrt{6}+\sqrt{2}}{4}}\right)}{2}} \\ \cos \frac{\pi }{24}=\sqrt{4+\sqrt{6}+\sqrt{2}} \\ \cos \frac{\pi }{24}=\frac{\sqrt{2}}{4}\sqrt{4+\sqrt{6}+\sqrt{2}} \end{gathered}$$