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 the product-to-sum identity.
We know that \[ \sin A\cos B=\frac{1}{2}\Big[\sin(A+B)+\sin(A-B)\Big] \] Taking \[ A=\frac{5\pi}{24}, \qquad B=\frac{\pi}{24}, \] we get \[ \sin\left(\frac{5\pi}{24}\right)\cos\left(\frac{\pi}{24}\right) = \frac{1}{2} \left[ \sin\left(\frac{6\pi}{24}\right) + \sin\left(\frac{4\pi}{24}\right) \right] \]

Step 2: Simplify the angles.
Since \[ \frac{6\pi}{24}=\frac{\pi}{4}, \] and \[ \frac{4\pi}{24}=\frac{\pi}{6}, \] the expression becomes \[ \frac{1}{2} \left[ \sin\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{6}\right) \right] \]

Step 3: Substitute the standard values.
We know that \[ \sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \] and \[ \sin\left(\frac{\pi}{6}\right)=\frac{1}{2} \] Therefore, \[ \frac{1}{2} \left[ \frac{\sqrt{2}}{2} + \frac{1}{2} \right] \] \[ = \frac{1}{2} \cdot \frac{\sqrt{2}+1}{2} \] \[ = \frac{\sqrt{2}+1}{4} \]

Step 4: Final conclusion.
Hence, \[ \sin\left(\frac{5\pi}{24}\right)\cos\left(\frac{\pi}{24}\right) = \boxed{\frac{1+\sqrt{2}}{4}} \] Thus, the correct option is \[ \boxed{(1)} \]