Question

$\cos 75^{\circ} \cos 45^{\circ} \cos 15^{\circ} = $ ________.

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

Hint:

$2\sin\theta\cos\theta = \sin(2\theta)$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
This problem involves evaluating a product of trigonometric functions. We can simplify the calculation by using trigonometric identities to combine terms.
Step 2: Key Formula or Approach
We will use the following trigonometric identities:
1. Co-function identity: \(\cos(90^\circ - \theta) = \sin(\theta)\)
2. Double-angle identity for sine: \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\), which can be rearranged to \(\sin(\theta)\cos(\theta) = \frac{1}{2}\sin(2\theta)\).
We also know the value of \(\cos(45^\circ) = \frac{1}{\sqrt{2}}\) and \(\sin(30^\circ) = \frac{1}{2}\).
Step 3: Detailed Explanation
1. Rearrange and simplify the expression.
The expression is \((\cos75^\circ \cos15^\circ) \cos45^\circ\).
Let's use the co-function identity on \(\cos75^\circ\).
\[ \cos75^\circ = \cos(90^\circ - 15^\circ) = \sin15^\circ \] Substituting this back into the expression, we get:
\[ (\sin15^\circ \cos15^\circ) \cos45^\circ \] 2. Apply the double-angle identity.
Using the identity \(\sin(\theta)\cos(\theta) = \frac{1}{2}\sin(2\theta)\) with \(\theta = 15^\circ\):
\[ \sin15^\circ \cos15^\circ = \frac{1}{2}\sin(2 \times 15^\circ) = \frac{1}{2}\sin(30^\circ) \] 3. Substitute known trigonometric values.
We know that \(\sin(30^\circ) = \frac{1}{2}\).
\[ \sin15^\circ \cos15^\circ = \frac{1}{2} \left(\frac{1}{2}\right) = \frac{1}{4} \] 4. Complete the calculation.
Now substitute this result back into the main expression:
\[ (\sin15^\circ \cos15^\circ) \cos45^\circ = \left(\frac{1}{4}\right) \cos45^\circ \] We also know that \(\cos(45^\circ) = \frac{1}{\sqrt{2}}\).
\[ \frac{1}{4} \times \frac{1}{\sqrt{2}} = \frac{1}{4\sqrt{2}} \] Step 4: Final Answer
The value of the expression is \(\frac{1}{4\sqrt{2}}\).