To prove the identity \(\cos 4\theta + 4\cos 2\theta + 3 \equiv 8\cos^4 \theta\), we start by expressing \(\cos 4\theta\) and \(\cos 2\theta\) in terms of \(\cos \theta\).

First, use the double angle formula for cosine:

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

Next, express \(\cos 4\theta\) using the double angle formula again:

\(\cos 4\theta = \cos(2 \times 2\theta) = 2\cos^2 2\theta - 1\)

Substitute \(\cos 2\theta = 2\cos^2 \theta - 1\) into the expression for \(\cos 4\theta\):

\(\cos 4\theta = 2(2\cos^2 \theta - 1)^2 - 1\)

Expand \((2\cos^2 \theta - 1)^2\):

\((2\cos^2 \theta - 1)^2 = 4\cos^4 \theta - 4\cos^2 \theta + 1\)

Substitute back into the expression for \(\cos 4\theta\):

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

Now substitute \(\cos 4\theta\) and \(\cos 2\theta\) into the original identity:

\(\cos 4\theta + 4\cos 2\theta + 3 = (8\cos^4 \theta - 8\cos^2 \theta + 1) + 4(2\cos^2 \theta - 1) + 3\)

Simplify the expression:

\(= 8\cos^4 \theta - 8\cos^2 \theta + 1 + 8\cos^2 \theta - 4 + 3\) \(= 8\cos^4 \theta\)

Thus, the identity is proven: \(\cos 4\theta + 4\cos 2\theta + 3 \equiv 8\cos^4 \theta\).

To prove the identity \(\cos 4\theta - 4 \cos 2\theta + 3 \equiv 8 \sin^4 \theta\), we start by using trigonometric identities.

First, express \(\cos 4\theta\) and \(\cos 2\theta\) using double angle formulas:

\(\cos 4\theta = 2\cos^2 2\theta - 1\) \(\cos 2\theta = 2\cos^2 \theta - 1\)

Substitute \(\cos 2\theta\) into the expression for \(\cos 4\theta\):

\(\cos 4\theta = 2(2\cos^2 \theta - 1)^2 - 1\)

Expand \((2\cos^2 \theta - 1)^2\):

\((2\cos^2 \theta - 1)^2 = 4\cos^4 \theta - 4\cos^2 \theta + 1\)

Substitute back:

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

Now substitute \(\cos 4\theta\) and \(\cos 2\theta\) into the left-hand side of the identity:

\(\cos 4\theta - 4\cos 2\theta + 3 = (8\cos^4 \theta - 8\cos^2 \theta + 1) - 4(2\cos^2 \theta - 1) + 3\)

Simplify:

\(= 8\cos^4 \theta - 8\cos^2 \theta + 1 - 8\cos^2 \theta + 4 + 3\) \(= 8\cos^4 \theta - 16\cos^2 \theta + 8\)

Now express in terms of \(\sin^2 \theta\):

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

Substitute:

\(8\cos^4 \theta - 16\cos^2 \theta + 8 = 8(1 - \sin^2 \theta)^2 - 16(1 - \sin^2 \theta) + 8\)

Expand:

\(= 8(1 - 2\sin^2 \theta + \sin^4 \theta) - 16 + 16\sin^2 \theta + 8\) \(= 8 - 16\sin^2 \theta + 8\sin^4 \theta - 16 + 16\sin^2 \theta + 8\) \(= 8\sin^4 \theta\)

Thus, the identity is proven.

To prove the identity \(\csc 2\theta + \cot 2\theta \equiv \cot \theta\), we start by expressing \(\csc\) and \(\cot\) in terms of sine and cosine.

We know:

\(\csc 2\theta = \frac{1}{\sin 2\theta}, \quad \cot 2\theta = \frac{\cos 2\theta}{\sin 2\theta}\)

Using the double angle formulas:

\(\sin 2\theta = 2 \sin \theta \cos \theta, \quad \cos 2\theta = \cos^2 \theta - \sin^2 \theta\)

Substitute these into the identity:

\(\csc 2\theta + \cot 2\theta = \frac{1}{2 \sin \theta \cos \theta} + \frac{\cos^2 \theta - \sin^2 \theta}{2 \sin \theta \cos \theta}\)

Combine the fractions:

\(= \frac{1 + \cos^2 \theta - \sin^2 \theta}{2 \sin \theta \cos \theta}\)

Since \(\cos^2 \theta - \sin^2 \theta = \cos 2\theta\), we have:

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

Using the identity \(1 + \cos 2\theta = 2 \cos^2 \theta\), we get:

\(= \frac{2 \cos^2 \theta}{2 \sin \theta \cos \theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta\)

Thus, the identity is proven.

To prove the identity \(\sin^2 \theta \cos^2 \theta \equiv \frac{1}{8}(1 - \cos 4\theta)\), we start by using the double angle identities.

First, recall the double angle identity for cosine:

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

We can express \(\sin^2 \theta \cos^2 \theta\) in terms of \(\cos 2\theta\):

\(\sin^2 \theta \cos^2 \theta = \left(\frac{1}{2} \sin 2\theta\right)^2 = \frac{1}{4} \sin^2 2\theta.\)

Using the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\), we have:

\(\sin^2 2\theta = \frac{1 - \cos 4\theta}{2}.\)

Substituting back, we get:

\(\sin^2 \theta \cos^2 \theta = \frac{1}{4} \cdot \frac{1 - \cos 4\theta}{2} = \frac{1}{8}(1 - \cos 4\theta).\)

Thus, the identity is proven.

To prove the identity \(\cot x - \cot 2x \equiv \csc 2x\), we start by using trigonometric identities.

Recall that \(\cot x = \frac{\cos x}{\sin x}\) and \(\cot 2x = \frac{\cos 2x}{\sin 2x}\).

Also, \(\csc 2x = \frac{1}{\sin 2x}\).

Using the double angle identities:

\(\sin 2x = 2 \sin x \cos x\)

\(\cos 2x = \cos^2 x - \sin^2 x\)

Now, substitute these into the expression:

\(\cot x - \cot 2x = \frac{\cos x}{\sin x} - \frac{\cos 2x}{\sin 2x}\)

\(= \frac{\cos x}{\sin x} - \frac{\cos^2 x - \sin^2 x}{2 \sin x \cos x}\)

Combine the fractions:

\(= \frac{2 \cos^2 x - (\cos^2 x - \sin^2 x)}{2 \sin x \cos x}\)

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

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

Using the Pythagorean identity \(\cos^2 x + \sin^2 x = 1\), we have:

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

\(= \frac{1}{\sin 2x}\)

\(= \csc 2x\)

Thus, the identity is proven: \(\cot x - \cot 2x \equiv \csc 2x\).