8 (a) Prove the identity $$\frac{\sin 2x}{1 + \tan^2 x} = 2 \sin x \cos^3 x$$ 8 (b) Hence find $$\int \frac{4 \sin 4\theta}{1 + \tan^2 2\theta} d\theta$$ - AQA - A-Level Maths Mechanics - Question 8 - 2018 - Paper 3

To prove the identity, we start by examining the left-hand side:

$\frac{\sin 2x}{1 + \tan^2 x}$

We can use the identity for ( \sin 2x = 2 \sin x \cos x ) to rewrite the numerator:

$\frac{2 \sin x \cos x}{1 + \tan^2 x}$

Next, using the identity ( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} ), we have:

$1 + \tan^2 x = 1 + \frac{\sin^2 x}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}$

Thus, we can rewrite the denominator as:

$\frac{\sin 2x}{\frac{1}{\cos^2 x}} = 2 \sin x \cos x \cdot \cos^2 x = 2 \sin x \cos^3 x$

This shows that:

$\frac{\sin 2x}{1 + \tan^2 x} = 2 \sin x \cos^3 x$

Now, for the right-hand side:

Since both sides are equal, we conclude that:

$\text{LHS} = \text{RHS}$