Question

$\sin 15^\circ =$

• A. $\frac{\sqrt{3}-1}{\sqrt{3}+2}$
• B. $\frac{\sqrt{6}-\sqrt{2}}{4}$
• C. $\sqrt{6} \pm 1$
• D. $\frac{\sqrt{6}+\sqrt{2}}{4}$

Hint:

Since $15^\circ$ is a small positive angle, its sine must be a small positive number. Note that $\sqrt{6} \approx 2.45$ and $\sqrt{2} \approx 1.41$, so $\frac{2.45-1.41}{4} \approx 0.26$, which is a logical value.

Answer 1

The Correct Option is B

1. Expressing the Angle: We can represent $15^\circ$ as the difference between two standard angles whose trigonometric values are well-known: $$15^\circ = 45^\circ - 30^\circ$$

2. Applying the Formula: The sine difference identity is $\sin(A - B) = \sin A \cos B - \cos A \sin B$. Let $A = 45^\circ$ and $B = 30^\circ$: $$\sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$$

3. Substituting Standard Values: Recall that: $\sin 45^\circ = \frac{1}{\sqrt{2}}$, $\cos 45^\circ = \frac{1}{\sqrt{2}}$ $\sin 30^\circ = \frac{1}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$ $$\sin 15^\circ = \left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)$$ $$\sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}-1}{2\sqrt{2}}$$