Question

1. prove:

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

Explanation:

Factor the numerator of the left-hand side

Using the Trigonometric Identities knowledge point

$$ 2\cos x + 2\cos^2 x = 2\cos x(1 + \cos x) $$

Apply the double-angle identity to the denominator

Using the Trigonometric Identities knowledge point

$$ \sin 2x = 2\sin x\cos x $$

Simplify the left-hand side expression

Using the Trigonometric Identities knowledge point

$$ \frac{2\cos x(1 + \cos x)}{2\sin x\cos x} = \frac{1 + \cos x}{\sin x} $$

Multiply by the conjugate to reach the target form

Using the Trigonometric Identities knowledge point

$$ \frac{1 + \cos x}{\sin x} \cdot \frac{1 - \cos x}{1 - \cos x} = \frac{1 - \cos^2 x}{\sin x(1 - \cos x)} = \frac{\sin^2 x}{\sin x(1 - \cos x)} = \frac{\sin x}{1 - \cos x} $$

Answer:

To prove the identity:

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

We simplify the left-hand side (LHS):

1. Factor the numerator:

$$ 2\cos x + 2\cos^2 x = 2\cos x(1 + \cos x) $$

1. Substitute the double-angle identity \(\sin 2x = 2\sin x\cos x\) into the denominator:

$$ \text{LHS} = \frac{2\cos x(1 + \cos x)}{2\sin x\cos x} $$

1. Cancel the common terms \(2\cos x\):

$$ \text{LHS} = \frac{1 + \cos x}{\sin x} $$

1. Multiply the numerator and denominator by the conjugate \(1 - \cos x\):

$$ \text{LHS} = \frac{(1 + \cos x)(1 - \cos x)}{\sin x(1 - \cos x)} = \frac{1 - \cos^2 x}{\sin x(1 - \cos x)} $$

1. Apply the Pythagorean identity \(1 - \cos^2 x = \sin^2 x\):

$$ \text{LHS} = \frac{\sin^2 x}{\sin x(1 - \cos x)} = \frac{\sin x}{1 - \cos x} $$

Since the simplified LHS is equal to the right-hand side (RHS), the identity is proven.