Chapter EP, Problem 14.3.6EP

To determine

To find: the exact value of the trigonometric expression.

Answer to Problem 14.3.6EP

  ${\text{sin15}}^0=\frac{\sqrt{\text{3}}\text{-1}}{\text{2}\sqrt{\text{2}}}$ .

Explanation of Solution

Given:

  ${\text{sin15}}^0$ .

Concept used:

The exact value of the trigonometric expression will be defined from the unit circle.

Some unit circle doesn’t provide enough value of angle so the angle should be taken reference to angle in the unit circle by taking difference or sum to make the appropriate angle which has value in the unit circle.

According to the trigonometric identities:

  $\text{sin}\left(\text{A-B}\right)\text{=sinAcosB-cosAsinB}$ .

Calculation:

  ${\text{sin15}}^0$ .

  $\Rightarrow {\text{sin15}}^0\text{=sin}{\left(\text{60-45}\right)}^{\text{0}}$ .

Since, the value of ${\text{60}}^0{\text{ and 45}}^0$ are in the unit circle.

  $\Rightarrow \text{sin}{\left(\text{60-45}\right)}^{\text{0}}$

According to the trigonometric identities:

  $\text{sin}\left(\text{A-B}\right)\text{=sinAcosB-cosAsinB}$ .

  $\text{sin}\left(\text{60-45}\right)\text{=sin60cos45-cos60sin45}$ .

From the unit circle the value of the angle can be deduced.

  $\Rightarrow \text{sin}\left(\text{60-45}\right)\text{=}\frac{\sqrt{\text{3}}}{\text{2}}\frac{\text{1}}{\sqrt{\text{2}}}\text{-}\frac{\text{1}}{\text{2}}\frac{\text{1}}{\sqrt{\text{2}}}$ .

  $\Rightarrow \text{sin}\left(\text{60-45}\right)\text{=}\frac{\sqrt{\text{3}}\text{-1}}{\text{2}\sqrt{\text{2}}}$ .

  $\Rightarrow \frac{\sqrt{\text{3}}\text{-1}}{\text{2}\sqrt{\text{2}}}$ .

Hence, ${\text{sin15}}^0=\frac{\sqrt{\text{3}}\text{-1}}{\text{2}\sqrt{\text{2}}}$ .