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) \(\frac{1}{4}\).

Answer 2

We are asked to find the value of: \(\sin\left(\frac{5\pi}{12}\right) \cdot \sin\left(\frac{\pi}{12}\right)\)

Use the identity: \[ \sin A \cdot \sin B = \frac{1}{2} \left[ \cos(A - B) - \cos(A + B) \right] \]

Let \( A = \frac{5\pi}{12} \), \( B = \frac{\pi}{12} \)

Then:
\[ \sin\left(\frac{5\pi}{12}\right) \cdot \sin\left(\frac{\pi}{12}\right) = \frac{1}{2} \left[ \cos\left(\frac{5\pi}{12} - \frac{\pi}{12}\right) - \cos\left(\frac{5\pi}{12} + \frac{\pi}{12}\right) \right] \]

Simplify:
\[ = \frac{1}{2} \left[ \cos\left(\frac{4\pi}{12}\right) - \cos\left(\frac{6\pi}{12}\right) \right] = \frac{1}{2} \left[ \cos\left(\frac{\pi}{3}\right) - \cos\left(\frac{\pi}{2}\right) \right] \]

We know: \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{2}\right) = 0\)

So: \[ = \frac{1}{2} \left( \frac{1}{2} - 0 \right) = \frac{1}{4} \]

Correct Answer: \(\frac{1}{4}\)

Answer 3

We want to find the value of the expression \(\sin \frac{5\pi}{12} \sin \frac{\pi}{12}\).v

We can use the product-to-sum trigonometric identity:

\[ \mathbf{2 \sin A \sin B = \cos(A-B) - \cos(A+B)} \]

or equivalently,

\[ \mathbf{\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]} \]

 

In this case, let \(A = \frac{5\pi}{12}\) and \(B = \frac{\pi}{12}\).

First, calculate \(A-B\) and \(A+B\):

\[ A - B = \frac{5\pi}{12} - \frac{\pi}{12} = \frac{4\pi}{12} = \mathbf{\frac{\pi}{3}} \]

\[ A + B = \frac{5\pi}{12} + \frac{\pi}{12} = \frac{6\pi}{12} = \mathbf{\frac{\pi}{2}} \]

Now, substitute these into the identity:

\[ \sin \frac{5\pi}{12} \sin \frac{\pi}{12} = \frac{1}{2} \left[ \cos\left(\frac{\pi}{3}\right) - \cos\left(\frac{\pi}{2}\right) \right] \]

We know the exact values of cosine for these angles:

• \(\cos\left(\frac{\pi}{3}\right) = \mathbf{\frac{1}{2}}\)
• \(\cos\left(\frac{\pi}{2}\right) = \mathbf{0}\)

Substitute these values back into the expression:

\[ \sin \frac{5\pi}{12} \sin \frac{\pi}{12} = \frac{1}{2} \left[ \frac{1}{2} - 0 \right] \]

\[ \sin \frac{5\pi}{12} \sin \frac{\pi}{12} = \frac{1}{2} \left[ \frac{1}{2} \right] \]

\[ \sin \frac{5\pi}{12} \sin \frac{\pi}{12} = \mathbf{\frac{1}{4}} \]

Comparing this with the given options, the correct option is:

\(\frac{1}{4}\)