Question

Find the value of: $ \sin 75^\circ \times \sin 15^\circ \times \sin 45^\circ $

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{8} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{1}{16} \)

Hint:

When dealing with trigonometric products, use known values and identities to simplify the calculation.

Answer 1

The Correct Option is B

We are tasked with evaluating: \[ \sin 75^\circ \times \sin 15^\circ \times \sin 45^\circ \] We can use the known values of these sine functions: - \( \sin 75^\circ = \sin (45^\circ + 30^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} \) - \( \sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4} \) - \( \sin 45^\circ = \frac{\sqrt{2}}{2} \) Now, calculate the product: \[ \sin 75^\circ \times \sin 15^\circ \times \sin 45^\circ = \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) \times \left( \frac{\sqrt{6} - \sqrt{2}}{4} \right) \times \frac{\sqrt{2}}{2} \] First, simplify the product of \( \sin 75^\circ \) and \( \sin 15^\circ \): \[ \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) \times \left( \frac{\sqrt{6} - \sqrt{2}}{4} \right) = \frac{(\sqrt{6})^2 - (\sqrt{2})^2}{16} = \frac{6 - 2}{16} = \frac{4}{16} = \frac{1}{4} \] Now, multiply by \( \sin 45^\circ \): \[ \frac{1}{4} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{8} \] 
Thus, the value is \( \frac{1}{8} \).