Question

The value of $\sin \frac{5\pi}{12} \sin \frac{\pi}{12}$ is

• A. 0
• B. $\frac{1}{2}$
• C. 1
• D. $\frac{1}{4}$

Answer 1

The Correct Option is D

We can use the trigonometric identity $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$
$\sin\left(\frac{5\pi}{12}\right) \sin\left(\frac{\pi}{12}\right) = \sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right) \sin\left(\frac{\pi}{12}\right)$
Using the identity, we have: 
$\sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right) \sin\left(\frac{\pi}{12}\right) = \left(\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{6}\right)\right) \sin\left(\frac{\pi}{12}\right)$

Simplifying further: 
$(\sin(\frac{\pi}{4})\cos(\frac{\pi}{6}) + \cos(\frac{\pi}{4})\sin(\frac{\pi}{6}))\sin(\frac{\pi}{12}) = (\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2})\sin(\frac{\pi}{12})$
Combining terms:
$\left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right)\sin\left(\frac{\pi}{12}\right) = \left(\frac{\sqrt{3} + 1}{2}\right) \cdot \sin\left(\frac{\pi}{12}\right)$

Now, $\sin\left(\frac{\pi}{12}\right)$ can be simplified using the half-angle formula: 
$\sin\left(\frac{\pi}{12}\right) = \sqrt{\frac{1 - \cos\left(\frac{\pi}{6}\right)}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$ 

Substituting back into the expression: 
$\left(\frac{\sqrt{3} + 1}{2}\right) \sin\left(\frac{\pi}{12}\right) = \left(\frac{\sqrt{3} + 1}{2}\right) \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$
Simplifying the expression:
 $\left(\frac{\sqrt{3} + 1}{2}\right) \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \left(\frac{\sqrt{3} + 1}{2}\right) \sqrt{\frac{2 - \sqrt{3}}{4}} = \left(\frac{\sqrt{3} + 1}{2}\right) \left(\frac{\sqrt{2} - \sqrt{3}}{2}\right)$

Multiplying the fractions:
$\left(\frac{\sqrt{3} + 1}{2}\right) \left(\frac{\sqrt{2} - \sqrt{3}}{2}\right) = \frac{\sqrt{6} - \sqrt{3} + \sqrt{2} - 1}{4}$
Therefore, the value of $\sin\left(\frac{5\pi}{12}\right) \sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{6} - \sqrt{3} + \sqrt{2} - 1}{4}$
Hence, the correct option is (D) $\frac{1}{4}$.