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CBSE · Class 11 · Mathematics · Trigonometric FunctionsWhat is the value of $\cos 15^{\circ}$?

Step-by-Step Solution

We write $\cos(15^{\circ}) = \cos(45^{\circ} - 30^{\circ})$. Using $\cos(A - B) = \cos A \cos B + \sin A \sin B$, we have $\cos 45^{\circ}\cos 30^{\circ} + \sin 45^{\circ}\sin 30^{\circ} = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}$.

Detailed Options Breakdown
Option : \frac{\sqrt{6} + \sqrt{2}}{4} (Correct Answer)

Correct choice. Refer to the step-by-step verified solution guidelines above for details.

Option 1: \frac{\sqrt{6} - \sqrt{2}}{4}

Incorrect choice. This distractor represents a common misunderstanding of the core principles of Trigonometric Functions.

Option 2: \frac{\sqrt{3} - 1}{4}

Incorrect choice. This distractor represents a common misunderstanding of the core principles of Trigonometric Functions.

Option 3: \frac{1}{2\sqrt{2}}

Incorrect choice. This distractor represents a common misunderstanding of the core principles of Trigonometric Functions.

💡 Study Guide: This question tests core syllabus concepts from Trigonometric Functions. For formulas, key summaries, and mock exam reference guides, read the full Trigonometric Functions Revision Notes.
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