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: \(\cot x - \cot 2x \equiv \csc 2x\).

To prove the identity \(\cot \theta - \tan \theta \equiv 2 \cot 2\theta\), we start by expressing each term in terms of sine and cosine.

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

Thus, the left-hand side (LHS) becomes:

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

Now, use the double-angle identity for cosine: \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\).

So, the LHS becomes:

\(\frac{\cos 2\theta}{\sin \theta \cos \theta}.\)

Now, consider the right-hand side (RHS):

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

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

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

Thus, the RHS is equal to the LHS, proving the identity:

\(\cot \theta - \tan \theta \equiv 2 \cot 2\theta.\)

To prove \(\csc 2\theta - \cot 2\theta \equiv \tan \theta\), we start by expressing the left-hand side in terms of \(\sin 2\theta\) and \(\cos 2\theta\).

We know:

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

Thus,

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

Using the double angle identities:

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

Substitute these into the expression:

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

Simplify the fraction:

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

Thus, \(\csc 2\theta - \cot 2\theta \equiv \tan \theta\) is proven.

To prove \(\frac{1 - \cos 2\theta}{1 + \cos 2\theta} \equiv \tan^2 \theta\), we start by using the double angle identity for cosine:

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

Substitute \(\cos 2\theta\) in the expression:

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

Simplify the expression:

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

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

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

Thus, \(\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \tan^2 \theta\) is proved.

To prove the identity \(\frac{2 \sin x - \sin 2x}{1 - \cos 2x} \equiv \frac{\sin x}{1 + \cos x}\), we start by simplifying the left-hand side (LHS).

First, use the double angle formula for sine: \(\sin 2x = 2 \sin x \cos x\).

Substitute \(\sin 2x\) in the LHS:

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

Simplify the numerator:

\(2 \sin x (1 - \cos x)\)

Now, use the double angle formula for cosine: \(\cos 2x = 1 - 2 \sin^2 x\).

Substitute \(\cos 2x\) in the denominator:

\(1 - (1 - 2 \sin^2 x) = 2 \sin^2 x\)

Thus, the LHS becomes:

\(\frac{2 \sin x (1 - \cos x)}{2 \sin^2 x}\)

Cancel \(2 \sin x\) from the numerator and denominator:

\(\frac{1 - \cos x}{\sin x}\)

Multiply by \(\frac{\sin x}{\sin x}\) to rationalize:

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

Thus, the identity is proven.

To prove the identity \(\frac{\cot x - \tan x}{\cot x + \tan x} \equiv \cos 2x\), we start by expressing \(\cot x\) and \(\tan x\) in terms of sine and cosine:

\(\cot x = \frac{\cos x}{\sin x}, \quad \tan x = \frac{\sin x}{\cos x}\)

Substitute these into the left-hand side (LHS):

\(\frac{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}}{\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}}\)

Find a common denominator for the fractions:

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

Simplify the expression:

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

Using the Pythagorean identity \(\cos^2 x + \sin^2 x = 1\), the expression simplifies to:

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

Recognize this as the double angle identity for cosine:

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

Thus, the identity is proven:

\(\frac{\cot x - \tan x}{\cot x + \tan x} \equiv \cos 2x\)

To prove the identity \(\tan 2\theta - \tan \theta \equiv \tan \theta \sec 2\theta\), we start by using the double angle formulas:

1. The double angle formula for tangent is:

\(\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}\)

2. The double angle formula for secant is:

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

3. Substitute \(\tan 2\theta\) into the left-hand side (LHS):

\(\tan 2\theta - \tan \theta = \frac{2\tan \theta}{1 - \tan^2 \theta} - \tan \theta\)

4. Combine into a single fraction:

\(\frac{2\tan \theta - \tan \theta (1 - \tan^2 \theta)}{1 - \tan^2 \theta} = \frac{2\tan \theta - \tan \theta + \tan^3 \theta}{1 - \tan^2 \theta}\)

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

5. Substitute \(\sec 2\theta\) into the right-hand side (RHS):

\(\tan \theta \sec 2\theta = \tan \theta \cdot \frac{2}{1 + \cos 2\theta}\)

\(= \frac{2\tan \theta}{1 - \tan^2 \theta}\)

6. Both sides are equal, thus proving the identity:

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

Therefore, the identity is verified.

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

1. Express \(\cos 4\theta\) using the double angle formula:

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

2. Express \(\cos 2\theta\) using the double angle formula:

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

3. Substitute \(\cos 2\theta\) into \(\cos 4\theta\):

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

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

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

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

4. Substitute into the left-hand side of the identity:

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

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

\(= 8\sin^4 \theta - 3\)

Thus, the identity is proven.

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

\(\cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta}\)

Thus,

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

Combine the fractions:

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

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

\(\frac{1}{\sin \theta \cos \theta}\)

Now, express \(\csc 2\theta\):

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

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

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

Therefore,

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

Thus, \(\cot \theta + \tan \theta = 2 \csc 2\theta\), proving the identity.

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

First, use the double angle formula for \(\cos 2\theta\):

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

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:

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

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

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

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

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

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

\(= 8\cos^4\theta - 3\)

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