Question

Evaluate \[ \cos^4 \frac{\pi}{24} - \sin^4 \frac{\pi}{24}. \]

• A. \(\frac{\sqrt{2}-\sqrt{3}}{2}\)
• B. \(\frac{\sqrt{2}+\sqrt{3}}{2}\)
• C. \(\frac{\sqrt{2}-\sqrt{6}}{4}\)
• D. \(\frac{\sqrt{2}+\sqrt{6}}{4}\)

Hint:

Use \(\cos^4 \theta - \sin^4 \theta = \cos 2\theta\) and express angles in sums/differences of known angles for exact trigonometric values.

Answer 1

The Correct Option is D

Step 1: Use difference of squares.
\[ \cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta) = \cos 2\theta \]

Step 2: Apply formula.
\[ \cos^4 \frac{\pi}{24} - \sin^4 \frac{\pi}{24} = \cos \frac{\pi}{12} \]

Step 3: Express \(\frac{\pi}{12}\) as difference.
\[ \frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6} \]

Step 4: Apply \(\cos(A-B)\) formula.
\[ \cos(A-B) = \cos A \cos B + \sin A \sin B \]

Step 5: Substitute values.
\[ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6} = \frac{1}{2} \]

Step 6: Compute final value.
\[ \cos \frac{\pi}{12} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4} \]
\[ \boxed{\frac{\sqrt{2}+\sqrt{6}}{4}} \]