Question

$\sin 15^{\circ} \sin 45^{\circ} \sin 75^{\circ} =$

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{4\sqrt{2}}$
• C. $\frac{1}{3\sqrt{2}}$
• D. $\frac{1}{4\sqrt{3}}$
• E. $\frac{1}{\sqrt{3}}$

Hint:

Using co-function identities like $\sin(90-x) = \cos x$ often simplifies trigonometric products.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to evaluate the product of three sine values. We can simplify the expression by using trigonometric identities, specifically the co-function identity and the double-angle formula.
Step 2: Key Formula or Approach:
1. Co-function identity: \( \sin(90^\circ - \theta) = \cos \theta \)
2. Double-angle identity for sine: \( \sin(2\theta) = 2 \sin \theta \cos \theta \), which can be rearranged to \( \sin \theta \cos \theta = \frac{1}{2} \sin(2\theta) \).
3. Standard trigonometric values: \( \sin 30^\circ = \frac{1}{2} \) and \( \sin 45^\circ = \frac{1}{\sqrt{2}} \).
Step 3: Detailed Explanation:
The given expression is \( \sin 15^\circ \sin 45^\circ \sin 75^\circ \).
First, let's simplify \( \sin 75^\circ \) using the co-function identity:
\[ \sin 75^\circ = \sin(90^\circ - 15^\circ) = \cos 15^\circ \] Now substitute this back into the original expression:
\[ \sin 15^\circ \sin 45^\circ \cos 15^\circ \] Rearrange the terms to group \( \sin 15^\circ \) and \( \cos 15^\circ \) together:
\[ (\sin 15^\circ \cos 15^\circ) \sin 45^\circ \] Apply the double-angle formula \( \sin \theta \cos \theta = \frac{1}{2} \sin(2\theta) \) with \( \theta = 15^\circ \):
\[ \left( \frac{1}{2} \sin(2 \cdot 15^\circ) \right) \sin 45^\circ = \frac{1}{2} \sin 30^\circ \sin 45^\circ \] Now, substitute the known values of \( \sin 30^\circ \) and \( \sin 45^\circ \):
\[ \frac{1}{2} \cdot \left( \frac{1}{2} \right) \cdot \left( \frac{1}{\sqrt{2}} \right) \] Multiply the terms:
\[ \frac{1}{4\sqrt{2}} \] Step 4: Final Answer:
The value of the expression is \( \frac{1}{4\sqrt{2}} \).