Question

The value of \( \sin(\frac{5\pi}{24}) \cdot \cos(\frac{\pi}{24}) \) is

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

Hint:

Product-to-Sum trigonometric identities:
\(2\sin A \cos B = \sin(A+B) + \sin(A-B)\)
\(2\cos A \sin B = \sin(A+B) - \sin(A-B)\)
\(2\cos A \cos B = \cos(A+B) + \cos(A-B)\)
\(2\sin A \sin B = \cos(A-B) - \cos(A+B)\)
Standard angle values: \(\sin(\pi/6)=1/2\), \(\sin(\pi/4)=\sqrt{2}/2\), \(\cos(\pi/6)=\sqrt{3}/2\), \(\cos(\pi/4)=\sqrt{2}/2\).

Answer 1

The Correct Option is A

We need to evaluate \( \sin(\frac{5\pi}{24}) \cos(\frac{\pi}{24}) \). Use the product-to-sum formula: \( \sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)] \). Let \(A = \frac{5\pi}{24}\) and \(B = \frac{\pi}{24}\). \(A+B = \frac{5\pi}{24} + \frac{\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}\). \(A-B = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6}\). So, \( \sin(\frac{5\pi}{24}) \cos(\frac{\pi}{24}) = \frac{1}{2}\left[\sin\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{6}\right)\right] \). We know the values: \(\sin(\pi/4) = \sin(45^\circ) = \frac{\sqrt{2}}{2}\) (or \(\frac{1}{\sqrt{2}}\)). \(\sin(\pi/6) = \sin(30^\circ) = \frac{1}{2}\). Substitute these values: \( = \frac{1}{2}\left[\frac{\sqrt{2}}{2} + \frac{1}{2}\right] = \frac{1}{2}\left[\frac{\sqrt{2}+1}{2}\right] = \frac{1+\sqrt{2}}{4} \). This matches option (a). \[ \boxed{\frac{1+\sqrt{2}}{4}} \]