Chapter 5, Problem 105RE

To determine

To calculate: The values of sine, cosine and tangent of the angle $15°$ .

Answer to Problem 105RE

The values are $\sin 15°=\frac{\sqrt{2-\sqrt{3}}}{2}$ , $\cos 15°=\frac{\sqrt{\sqrt{3}-2}}{2}$ , $\tan 15°=\frac{\sqrt{2-\sqrt{3}}}{\sqrt{\sqrt{3}-2}}$ .

Explanation of Solution

Given Information:

The angle is $15°$ .

Formula Used

The trigonometric formulas are ${\cos }^2\theta =\frac{\cos 2\theta -1}{2}$ , ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$ and $\tan x=\frac{\sin x}{\cos x}$ .

Calculation:

The given angle is $15°$ .

Rewrite it as $15°=\frac{30°}{2}$ .

Know that ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$ .

Substitute $\frac{30°}{2}$ for $\theta$ .

  $\begin{array}{c}{\sin }^2\frac{30°}{2}=\frac{1-\cos 30°}{2}\\ =\frac{1-\frac{\sqrt{3}}{2}}{2}\\ =\frac{2-\sqrt{3}}{4}\end{array}$

Take square root in both side.

  $\begin{array}{c}\sqrt{{\sin }^2\frac{30°}{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}\\ \sin \frac{30°}{2}=\sqrt{\frac{2-\sqrt{3}}{4}}\\ \sin 15°=\frac{\sqrt{2-\sqrt{3}}}{2}\end{array}$

Know that, ${\cos }^2\theta =\frac{\cos 2\theta -1}{2}$ .

  $\begin{array}{c}{\cos }^2\frac{30°}{2}=\frac{\cos 30°-1}{2}\\ =\frac{\frac{\sqrt{3}}{2}-1}{2}\\ =\frac{\sqrt{3}-2}{4}\end{array}$

Take square root in both side.

  $\begin{array}{c}\sqrt{{\cos }^2\frac{30°}{2}}=\sqrt{\frac{\sqrt{3}-2}{4}}\\ \cos \frac{30°}{2}=\sqrt{\frac{\sqrt{3}-2}{4}}\\ \cos 15°=\frac{\sqrt{\sqrt{3}-2}}{2}\end{array}$

Know that $\tan \theta =\frac{\sin \theta }{\cos \theta }$ .

  $\begin{array}{c}\tan \frac{30°}{2}=\tan 15°\\ =\frac{\sin 15°}{\cos 15°}\\ =\frac{\frac{\sqrt{2-\sqrt{3}}}{2}}{\frac{\sqrt{\sqrt{3}-2}}{2}}\\ =\frac{\sqrt{2-\sqrt{3}}}{\sqrt{\sqrt{3}-2}}\end{array}$

Hence, the values are $\sin 15°=\frac{\sqrt{2-\sqrt{3}}}{2}$ , $\cos 15°=\frac{\sqrt{\sqrt{3}-2}}{2}$ , $\tan 15°=\frac{\sqrt{2-\sqrt{3}}}{\sqrt{\sqrt{3}-2}}$ .