Mathematics — Standard (041) — CBSE Class 10 board question

Write $15° = 45° - 30°$, so $\sin 15° = \sin(45° - 30°)$.

Using the given identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:

$$\sin 15° = \sin 45°\cos 30° - \cos 45°\sin 30°$$

$$= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2}$$

$$= \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}-1}{2\sqrt{2}}$$

Source: Chapter 8, Section 8.3 (Table 8.1)

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• The key step is recognising $15° = 45° - 30°$ so the given formula can be applied directly with standard angles from Table 8.1.
• Examiners award 1 mark for correct substitution of standard values and 1 mark for the simplified final answer $\dfrac{\sqrt{3}-1}{2\sqrt{2}}$.
• You may also write this as $\dfrac{\sqrt{6}-\sqrt{2}}{4}$ (rationalised form) — both are acceptable.