Solving the Trigonometric Equation

The problem asks for the value of the expression \( \cos A \cos(A/2) \sin(A/2) \) given the equation \( 2\sqrt{3} \sin A \cos A = \cos 2A – \sin 2A \), and that \( A \) lies in the first quadrant.

First, let's simplify the given trigonometric equation:

\( 2\sqrt{3} \sin A \cos A = \cos 2A – \sin 2A \)

We can use the double angle identities:

• \( \sin 2A = 2 \sin A \cos A \)
• \( \cos 2A = \cos^2 A – \sin^2 A \)

Substitute \( \sin 2A \) into the left side of the equation:

\( \sqrt{3} (2 \sin A \cos A) = \sqrt{3} \sin 2A \)

Substitute the double angle identity for \( \cos 2A \) into the right side is not necessary here. The right side is already in terms of \( \cos 2A \).

So, the equation becomes:

\( \sqrt{3} \sin 2A = \cos 2A \)

To solve for \( 2A \), we can divide both sides by \( \cos 2A \) (assuming \( \cos 2A \neq 0 \)):

\( \frac{\sqrt{3} \sin 2A}{\cos 2A} = \frac{\cos 2A}{\cos 2A} \)

\( \sqrt{3} \tan 2A = 1 \)

\( \tan 2A = \frac{1}{\sqrt{3}} \)

We are given that \( A \) lies in the first quadrant, which means \( 0 < A < \frac{\pi}{2} \) radians or \( 0^\circ < A < 90^\circ \).

Multiplying the inequality by 2, we get \( 0 < 2A < \pi \) radians or \( 0^\circ < 2A < 180^\circ \).

The general solutions for \( \tan x = \frac{1}{\sqrt{3}} \) are \( x = n\pi + \frac{\pi}{6} \), where \( n \) is an integer.

For \( 0 < 2A < \pi \), the only possible value for \( 2A \) is when \( n=0 \):

\( 2A = \frac{\pi}{6} \)

From this, we find the value of \( A \):

\( A = \frac{\pi}{12} \)

This value \( \frac{\pi}{12} \) (or \( 15^\circ \)) is indeed in the first quadrant (\( 0 < \frac{\pi}{12} < \frac{\pi}{2} \)).

Evaluating the Expression \( \cos A \cos(A/2) \sin(A/2) \)

Now we need to find the value of \( \cos A \cos(A/2) \sin(A/2) \). Let's simplify this expression.

Recall the double angle identity for sine: \( \sin x = 2 \sin(x/2) \cos(x/2) \). This can be rearranged as \( \sin(x/2) \cos(x/2) = \frac{1}{2} \sin x \).

Applying this to the \( \cos(A/2) \sin(A/2) \) part of our expression, with \( x=A \), we get:

\( \cos(A/2) \sin(A/2) = \frac{1}{2} \sin A \)

Substitute this back into the expression we want to evaluate:

\( \cos A \left( \frac{1}{2} \sin A \right) = \frac{1}{2} \sin A \cos A \)

We can use another double angle identity: \( \sin 2A = 2 \sin A \cos A \), which means \( \sin A \cos A = \frac{1}{2} \sin 2A \).

Substitute this into the simplified expression:

\( \frac{1}{2} \left( \frac{1}{2} \sin 2A \right) = \frac{1}{4} \sin 2A \)

Now, we substitute the value of \( 2A \) we found from the equation, which is \( 2A = \frac{\pi}{6} \).

\( \frac{1}{4} \sin \left( \frac{\pi}{6} \right) \)

We know that \( \sin \left( \frac{\pi}{6} \right) = \sin(30^\circ) = \frac{1}{2} \).

So, the value of the expression is:

\( \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \)

Final Value of the Expression

The value of \( \cos A \cos(A/2) \sin(A/2) \) is \( \frac{1}{8} \).

Let's verify the options provided:

Our calculated value matches Option 2.