Question

Show that$\cos \frac{\pi }{8}=\frac{\sqrt{2+\sqrt{2}}}{2}$and use it to find$\sin \frac{\pi }{16},\cos \frac{\pi }{16}$.

Short answer

The value of $\sin \frac{\pi }{16}$is $\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}$.

The value of $\cos \frac{\pi }{16}$is $\frac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2}$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\cos \frac{\pi }{8}=\frac{\sqrt{2+\sqrt{2}}}{2}$

Using half-angle formulas,

$$\begin{gathered} \cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}} \\ \sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}} \end{gathered}$$

Consider $\cos \theta =\cos \left(\frac{\mathrm{\pi }}{4}\right)$. Then,

localid="1646421590598" $$\begin{gathered} \theta =\frac{\mathrm{\pi }}{4} \\ \frac{\theta }{2}=\frac{\mathrm{\pi }}{8} \end{gathered}$$

Step 2. Use the half angle formula.

We know that $\cos \left(\frac{\mathrm{\pi }}{4}\right)=\frac{\sqrt{2}}{2}$.

Using half-angle formulas,

$$\begin{gathered} \cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}} \\ \cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\left({\displaystyle \frac{\sqrt{2}}{2}}\right)}{2}} \\ \cos \frac{\theta }{2}=\pm \sqrt{\frac{2+\sqrt{2}}{2}} \end{gathered}$$

As $\frac{\theta }{2}$lies in the first quadrant,

$\cos \left(\frac{\mathrm{\pi }}{8}\right)=\frac{\sqrt{2+\sqrt{2}}}{2}$

Step 3. Use the trigonometric identities.

Consider $\cos \theta =\cos \left(\frac{\mathrm{\pi }}{8}\right)$. Then,

role="math" localid="1646421897677" $$\begin{gathered} \theta =\frac{\mathrm{\pi }}{8} \\ \frac{\theta }{2}=\frac{\mathrm{\pi }}{16} \end{gathered}$$

As $\frac{\theta }{2}$lies in the first quadrant,

$\cos \left(\frac{\mathrm{\pi }}{8}\right)=\frac{\sqrt{2+\sqrt{2}}}{2}$

Using half-angle formulas,

$$\begin{gathered} \cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}} \\ \cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}} \\ \cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\left({\displaystyle \frac{\sqrt{2+\sqrt{2}}}{2}}\right)}{2}} \\ \cos \frac{\theta }{2}=\pm \sqrt{\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}} \end{gathered}$$

Step 4. Solve for sinπ16.

Consider the given question,

$$\begin{gathered} \sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}} \\ \sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\left({\displaystyle \frac{\sqrt{2+\sqrt{2}}}{2}}\right)}{2}} \\ \sin \frac{\theta }{2}=\pm \frac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2} \end{gathered}$$