Evaluating the Trigonometric Expression 4cos(\(\frac{\pi}{6}\, - \,\alpha\)) sin(\(\frac{\pi}{3}\, - \,\alpha\))

The question asks us to find the value of the expression \(4\cos\left( {\frac{\pi }{6}\, - \,\alpha } \right)\) sin\(\left( {\frac{\pi }{3}\, - \,\alpha } \right)\). We can simplify this expression using trigonometric identities, specifically the product-to-sum identity.

The relevant product-to-sum identity is:

\(2 \cos A \sin B = \sin(A+B) - \sin(A-B)\)

Let's rewrite the given expression slightly to match the identity format:

\(4\cos\left( {\frac{\pi }{6}\, - \,\alpha } \right) \sin\left( {\frac{\pi }{3}\, - \,\alpha } \right) = 2 \times \left[ 2 \cos\left( {\frac{\pi }{6}\, - \,\alpha } \right) \sin\left( {\frac{\pi }{3}\, - \,\alpha } \right) \right]\)

Now, let \(A = \frac{\pi}{3} - \alpha\) and \(B = \frac{\pi}{6} - \alpha\). Applying the product-to-sum identity to the term inside the square brackets, we get \(2 \cos B \sin A = \sin(A+B) - \sin(B-A)\).

First, calculate \(A+B\):

\(A+B = \left( \frac{\pi}{3} - \alpha \right) + \left( \frac{\pi}{6} - \alpha \right) = \frac{\pi}{3} + \frac{\pi}{6} - \alpha - \alpha = \frac{2\pi + \pi}{6} - 2\alpha = \frac{3\pi}{6} - 2\alpha = \frac{\pi}{2} - 2\alpha\)

Next, calculate \(B-A\):

\(B-A = \left( \frac{\pi}{6} - \alpha \right) - \left( \frac{\pi}{3} - \alpha \right) = \frac{\pi}{6} - \alpha - \frac{\pi}{3} + \alpha = \frac{\pi}{6} - \frac{\pi}{3} = \frac{\pi - 2\pi}{6} = -\frac{\pi}{6}\)

Now, substitute these values into the identity:

\(2 \cos\left( {\frac{\pi }{6}\, - \,\alpha } \right) \sin\left( {\frac{\pi }{3}\, - \,\alpha } \right) = \sin\left( \frac{\pi}{2} - 2\alpha \right) - \sin\left( -\frac{\pi}{6} \right)\)

Using the identities \(\sin(\frac{\pi}{2} - \theta) = \cos \theta\) and \(\sin(-\theta) = -\sin \theta\):

• \(\sin\left( \frac{\pi}{2} - 2\alpha \right) = \cos(2\alpha)\)
• \(\sin\left( -\frac{\pi}{6} \right) = -\sin\left( \frac{\pi}{6} \right)\)

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

So, \(2 \cos\left( {\frac{\pi }{6}\, - \,\alpha } \right) \sin\left( {\frac{\pi }{3}\, - \,\alpha } \right) = \cos(2\alpha) - \left(-\frac{1}{2}\right) = \cos(2\alpha) + \frac{1}{2}\)

Now, substitute this back into the original expression, which was \(2 \times \left[ 2 \cos\left( {\frac{\pi }{6}\, - \,\alpha } \right) \sin\left( {\frac{\pi }{3}\, - \,\alpha } \right) \right]\):

\(4\cos\left( {\frac{\pi }{6}\, - \,\alpha } \right) \sin\left( {\frac{\pi }{3}\, - \,\alpha } \right) = 2 \times \left( \cos(2\alpha) + \frac{1}{2} \right) = 2\cos(2\alpha) + 1\)

The options are given in terms of \(\sin^2 \alpha\). We can use the double angle identity for cosine: \(\cos(2\alpha) = 1 - 2\sin^2 \alpha\).

Substitute this into the result:

\(2\cos(2\alpha) + 1 = 2(1 - 2\sin^2 \alpha) + 1 = 2 - 4\sin^2 \alpha + 1 = 3 - 4\sin^2 \alpha\)

Thus, the value of the expression is \(3 - 4\sin^2 \alpha\).

Let's compare this with the given options:

1. \(3 + 4\sin^2 \alpha\)
2. \(3 + \sin^2 \alpha\)
3. \(3 - \sin^2 \alpha\)
4. \(3 - 4\sin^2 \alpha\)

Our calculated value matches option 4.

Revision Table: Key Trigonometric Identities

Additional Information: Working with Product-to-Sum Identities

Product-to-sum identities are useful for converting products of sines and cosines into sums or differences. This transformation is often necessary in integration, or when simplifying complex trigonometric expressions like the one in this question. The main product-to-sum identities are:

• \(2 \sin A \cos B = \sin(A+B) + \sin(A-B)\)
• \(2 \cos A \sin B = \sin(A+B) - \sin(A-B)\) (Used in this problem)
• \(2 \cos A \cos B = \cos(A+B) + \cos(A-B)\)
• \(2 \sin A \sin B = \cos(A-B) - \cos(A+B)\)

These identities are derived from the sum and difference formulas for sine and cosine. For example, adding the sum and difference formulas for sine gives:

\(\sin(A+B) = \sin A \cos B + \cos A \sin B\)

\(\sin(A-B) = \sin A \cos B - \cos A \sin B\)

Adding these two equations yields:

\(\sin(A+B) + \sin(A-B) = 2 \sin A \cos B\)

Subtracting the second from the first yields:

\(\sin(A+B) - \sin(A-B) = 2 \cos A \sin B\)

Similar derivations lead to the other identities. Being familiar with these transformations is crucial for solving various problems in trigonometry and calculus.