Given that cos A = \frac{\sqrt{3}}{4}, where 270° < A < 360°, find the exact value of sin 2A - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 4

To prove this statement, we can use the cosine addition formula:

$\cos(a + b) + \cos(a - b) = 2 \cos a \cos b.$

In this case, let:

• (a = 2x)
• (b = \frac{\pi}{3})

Then:

$\cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - \frac{\pi}{3} \right) = 2 \cos \left(2x\right) \cos \left(\frac{\pi}{3}\right).$

Knowing that (\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}), the equation simplifies to:

$2 \cos(2x) \cdot \frac{1}{2} = \cos(2x).$

This confirms that:

$\cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - \frac{\pi}{3} \right) = \cos 2x.$

Given:

$y = 3 \sin^2 x + \cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - \frac{\pi}{3} \right).$

Using the chain rule for the first term and the result from part (b)(i) for the second terms:

1. The derivative of the first term: $\frac{d}{dx} [3 \sin^2 x] = 3 \cdot 2 \sin x \cos x = 6 \sin x \cos x,$ which can be rewritten as:
   $3 \sin 2x.$

2. From part (b)(i): $\cos \left( 2x + \frac{\pi}{3} \right) + \cos \left( 2x - \frac{\pi}{3} \right) = \cos 2x.$
   Thus, the derivative will be(-\sin 2x \cdot 2 = -2 \sin 2x.$$

Combining these together:

$\frac{dy}{dx} = 3 \sin 2x - 2 \sin 2x = \sin 2x.$

Therefore, we have shown that:

$\frac{dy}{dx} = \sin 2x.$