Question

The value of \[ \sin\left(\frac{5\pi}{24}\right)\cos\left(\frac{\pi}{24}\right) \] is

• A. \(\dfrac{1+\sqrt{2}}{4}\)
• B. \(1+\sqrt{2}\)
• C. \(\dfrac{1-\sqrt{2}}{4}\)
• D. \(1-\sqrt{2}\)

Hint:

Whenever a product of \(\sin\) and \(\cos\) appears, try using the identity \[ \sin A\cos B=\frac{1}{2}\big[\sin(A+B)+\sin(A-B)\big]. \] It often converts the expression into standard trigonometric values.

Answer 1

The Correct Option is A

Step 1: Use product-to-sum.
\[\sin\frac{5\pi}{24}\cos\frac{\pi}{24} = \frac{1}{2}\left[\sin\frac{6\pi}{24}+\sin\frac{4\pi}{24}\right] = \frac{1}{2}\left[\sin\frac{\pi}{4}+\sin\frac{\pi}{6}\right].\]

Step 2: Substitute values.
\[= \frac{1}{2}\left[\frac{1}{\sqrt{2}}+\frac{1}{2}\right] = \frac{1}{2}\cdot\frac{2+\sqrt{2}}{2\sqrt{2}} \cdot\sqrt{2} = \frac{1}{2}\cdot\frac{\sqrt{2}+1}{2} \cdot \frac{\sqrt{2}}{\sqrt{2}} \] Directly: \(\tfrac{1}{2}(\tfrac{\sqrt{2}}{2}+\tfrac{1}{2}) = \tfrac{1}{2}\cdot\tfrac{\sqrt{2}+1}{2} = \tfrac{1+\sqrt{2}}{4}\).
\[\boxed{\frac{1+\sqrt{2}}{4}}\]