(i) Start with the identity for \(\cot \theta\) and \(\tan \theta\):

\(\cot \theta = \frac{\cos \theta}{\sin \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

Add them: \(\cot \theta + \tan \theta = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}\).

Using the double angle identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have:

\(\frac{1}{\sin \theta \cos \theta} = \frac{2}{2 \sin \theta \cos \theta} = \frac{2}{\sin 2\theta} = 2 \csc 2\theta\).

Thus, \(\cot \theta + \tan \theta \equiv 2 \csc 2\theta\).

(ii) We need to evaluate \(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \csc 2\theta \, d\theta\).

Using the identity \(\csc 2\theta = \frac{1}{\sin 2\theta}\), the integral becomes:

\(\int \csc 2\theta \, d\theta = \int \frac{1}{\sin 2\theta} \, d\theta\).

Let \(u = 2\theta\), then \(du = 2 \, d\theta\) or \(d\theta = \frac{1}{2} \, du\).

The integral becomes \(\frac{1}{2} \int \csc u \, du\).

The integral of \(\csc u\) is \(\ln |\csc u - \cot u|\), so:

\(\frac{1}{2} \ln |\csc u - \cot u|\).

Substitute back \(u = 2\theta\):

\(\frac{1}{2} \ln |\csc 2\theta - \cot 2\theta|\).

Evaluate from \(\theta = \frac{1}{6}\pi\) to \(\theta = \frac{1}{3}\pi\):

\(\frac{1}{2} \left[ \ln |\csc \frac{2}{3}\pi - \cot \frac{2}{3}\pi| - \ln |\csc \frac{1}{3}\pi - \cot \frac{1}{3}\pi| \right]\).

Calculate the values:

\(\csc \frac{2}{3}\pi = \frac{1}{\sin \frac{2}{3}\pi} = \frac{2}{\sqrt{3}}\), \(\cot \frac{2}{3}\pi = \frac{1}{\tan \frac{2}{3}\pi} = -\frac{1}{\sqrt{3}}\).

\(\csc \frac{1}{3}\pi = \frac{1}{\sin \frac{1}{3}\pi} = 2\), \(\cot \frac{1}{3}\pi = \frac{1}{\tan \frac{1}{3}\pi} = \sqrt{3}\).

Substitute these back:

\(\frac{1}{2} \left[ \ln \left| \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} \right| - \ln \left| 2 - \sqrt{3} \right| \right]\).

\(\frac{1}{2} \left[ \ln \left( \frac{3}{\sqrt{3}} \right) - \ln \left( 2 - \sqrt{3} \right) \right]\).

\(\frac{1}{2} \left[ \ln 3 - \ln 1 \right] = \frac{1}{2} \ln 3\).

(i) Express \(\cos 4\theta\) as \(2 \cos^2 2\theta - 1\) or \(\cos^2 2\theta - \sin^2 2\theta\) or \(1 - 2 \sin^2 2\theta\).

Express \(\cos 4\theta\) in terms of \(\cos \theta\).

Obtain \(8 \cos^4 \theta - 8 \cos^2 \theta + 1\).

Use \(\cos 2\theta = 2 \cos^2 \theta - 1\) to obtain given answer \(8 \cos^4 \theta - 3\).

(ii) (a) State or imply \(\cos^4 \theta = \frac{1}{2}\).

Obtain \(0.572\).

Obtain \(-0.572\).

(b) Integrate and obtain form \(k_1 \theta + k_2 \sin 4\theta + k_3 \sin 2\theta\).

Obtain \(\frac{3}{8} \theta + \frac{1}{32} \sin 4\theta + \frac{1}{4} \sin 2\theta\).

Obtain \(\frac{3}{32}\pi + \frac{1}{4}\) following completely correct work.

(i) To find \(f'(x)\), we first differentiate \(f(x) = 4 \cos^2 3x\) using the chain rule. Let \(u = \cos 3x\), then \(f(x) = 4u^2\). The derivative \(\frac{du}{dx} = -3 \sin 3x\).

Using the chain rule, \(\frac{df}{dx} = 8u \cdot \frac{du}{dx} = 8 \cos 3x (-3 \sin 3x) = -24 \cos 3x \sin 3x\).

Now, evaluate \(f'(\frac{1}{9}\pi)\):

\(\cos 3(\frac{1}{9}\pi) = \cos \frac{\pi}{3} = \frac{1}{2}\)

\(\sin 3(\frac{1}{9}\pi) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\)

Thus, \(f'(\frac{1}{9}\pi) = -24 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = -6\sqrt{3}\).

(ii) To integrate \(f(x) = 4 \cos^2 3x\), use the identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\).

\(f(x) = 4 \cdot \frac{1 + \cos 6x}{2} = 2 + 2 \cos 6x\).

Integrate term by term:

\(\int (2 + 2 \cos 6x) \, dx = \int 2 \, dx + \int 2 \cos 6x \, dx\)

\(= 2x + \frac{2}{6} \sin 6x + C = 2x + \frac{1}{3} \sin 6x + C\).

(i) To prove the identity \(\cos 3\theta \equiv 4 \cos^3 \theta - 3 \cos \theta\), we start with the triple angle formula:

\(\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta\).

Using \(\cos 2\theta = 2\cos^2 \theta - 1\) and \(\sin 2\theta = 2\sin \theta \cos \theta\), we have:

\(\cos 3\theta = (2\cos^2 \theta - 1)\cos \theta - (2\sin \theta \cos \theta)\sin \theta\).

\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2\sin^2 \theta \cos \theta\).

Using \(\sin^2 \theta = 1 - \cos^2 \theta\), we get:

\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2(1 - \cos^2 \theta)\cos \theta\).

\(\cos 3\theta = 2\cos^3 \theta - \cos \theta - 2\cos \theta + 2\cos^3 \theta\).

\(\cos 3\theta = 4\cos^3 \theta - 3\cos \theta\).

(ii) Using the identity, we find the integral:

\(\int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \cos^3 \theta \, d\theta = \int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \frac{1}{4}(\cos 3\theta + 3\cos \theta) \, d\theta\).

\(= \frac{1}{4} \left[ \int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \cos 3\theta \, d\theta + 3 \int_{\frac{1}{3}\pi}^{\frac{1}{2}\pi} \cos \theta \, d\theta \right]\).

The integral of \(\cos 3\theta\) is \(\frac{1}{3} \sin 3\theta\), and the integral of \(\cos \theta\) is \(\sin \theta\).

\(= \frac{1}{4} \left[ \frac{1}{3} (\sin(3 \times \frac{1}{2}\pi) - \sin(3 \times \frac{1}{3}\pi)) + 3 (\sin(\frac{1}{2}\pi) - \sin(\frac{1}{3}\pi)) \right]\).

\(= \frac{1}{4} \left[ \frac{1}{3} (0 - \frac{\sqrt{3}}{2}) + 3 (1 - \frac{1}{2}) \right]\).

\(= \frac{1}{4} \left[ -\frac{\sqrt{3}}{6} + \frac{3}{2} \right]\).

\(= \frac{1}{4} \left[ \frac{3}{2} - \frac{\sqrt{3}}{6} \right]\).

\(= \frac{1}{4} \left[ \frac{9}{6} - \frac{\sqrt{3}}{6} \right]\).

\(= \frac{1}{4} \times \frac{9 - \sqrt{3}}{6}\).

\(= \frac{9 - \sqrt{3}}{24}\).

\(= \frac{2}{3} - \frac{3}{8} \sqrt{3}\).

(i) Use the angle subtraction and addition formulas:

\(\cos(3x-x) = \cos 2x = \cos 3x \cos x + \sin 3x \sin x\)

\(\cos(3x+x) = \cos 4x = \cos 3x \cos x - \sin 3x \sin x\)

Subtract these equations:

\(\cos 2x - \cos 4x = 2 \sin 3x \sin x\)

Thus, \(\frac{1}{2}(\cos 2x - \cos 4x) = \sin 3x \sin x\).

(ii) The integral becomes:

\(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \sin 3x \sin x \, dx = \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \left( \frac{1}{4} \sin 2x - \frac{1}{8} \sin 4x \right) \, dx\)

Integrate each term:

\(\int \sin 2x \, dx = -\frac{1}{2} \cos 2x\)

\(\int \sin 4x \, dx = -\frac{1}{4} \cos 4x\)

Evaluate from \(\frac{1}{6}\pi\) to \(\frac{1}{3}\pi\):

\(\left[ -\frac{1}{8} \cos 2x + \frac{1}{32} \cos 4x \right]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi}\)

Calculate the definite integral:

\(= \left( -\frac{1}{8} \cos \frac{2}{3}\pi + \frac{1}{32} \cos \frac{4}{3}\pi \right) - \left( -\frac{1}{8} \cos \frac{1}{3}\pi + \frac{1}{32} \cos \frac{2}{3}\pi \right)\)

\(= \frac{1}{8}\sqrt{3}\)