Past Exam Questions

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: \(\cos 4\theta + 4\cos 2\theta \equiv 8\cos^4\theta - 3\).

(a) To express \(3 \cos x + 2 \cos(x - 60^\circ)\) in the form \(R \cos(x - \alpha)\):

First, expand \(2 \cos(x - 60^\circ)\):

\(2 \cos(x - 60^\circ) = 2(\cos x \cos 60^\circ + \sin x \sin 60^\circ) = \cos x + \sqrt{3} \sin x\)

Combine with \(3 \cos x\):

\(3 \cos x + \cos x + \sqrt{3} \sin x = 4 \cos x + \sqrt{3} \sin x\)

Now, use the identity:

\(a \cos \theta + b \sin \theta \equiv R \cos(\theta - \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 4\) and \(b = \sqrt{3}\):

\(R = \sqrt{4^2 + (\sqrt{3})^2} = \sqrt{16 + 3} = \sqrt{19}\) \(\tan \alpha = \frac{\sqrt{3}}{4}\)

Thus, \(\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \approx 23.41^\circ\).

(b) Solve \(3 \cos 2\theta + 2 \cos(2\theta - 60^\circ) = 2.5\):

Using the result from part (a), express:

\(3 \cos 2\theta + 2 \cos(2\theta - 60^\circ) = \sqrt{19} \cos(2\theta - 23.41^\circ)\)

Set equal to 2.5:

\(\sqrt{19} \cos(2\theta - 23.41^\circ) = 2.5\)

\(\cos(2\theta - 23.41^\circ) = \frac{2.5}{\sqrt{19}}\)

Find \(2\theta\):

\(2\theta - 23.41^\circ = \cos^{-1}\left(\frac{2.5}{\sqrt{19}}\right)\)

Calculate \(2\theta\):

\(2\theta = \cos^{-1}\left(\frac{2.5}{\sqrt{19}}\right) + 23.41^\circ\)

\(\theta = 39.2^\circ\) and \(\theta = 82.1^\circ\) (considering the range \(0^\circ < \theta < 180^\circ\)).

(i) To express \(\sqrt{6} \sin x + \cos x\) in the form \(R \sin(x + \alpha)\), we use the identity:

\(a \sin x + b \cos x \equiv R \sin(x + \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = \sqrt{6}\) and \(b = 1\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{6})^2 + 1^2} = \sqrt{6 + 1} = \sqrt{7}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{1}{\sqrt{6}}\)

\(\alpha = \tan^{-1}\left(\frac{1}{\sqrt{6}}\right) \approx 22.208^\circ\)

(ii) Solve \(\sqrt{6} \sin 2\theta + \cos 2\theta = 2\).

Express in the form \(R \sin(2\theta + \alpha)\):

\(\sqrt{6} \sin 2\theta + \cos 2\theta = \sqrt{7} \sin(2\theta + 22.208^\circ)\)

Set equal to 2:

\(\sqrt{7} \sin(2\theta + 22.208^\circ) = 2\)

\(\sin(2\theta + 22.208^\circ) = \frac{2}{\sqrt{7}}\)

Calculate \(2\theta + 22.208^\circ\):

\(2\theta + 22.208^\circ = \sin^{-1}\left(\frac{2}{\sqrt{7}}\right) \approx 49.107^\circ\)

\(2\theta = 49.107^\circ - 22.208^\circ\)

\(2\theta = 26.899^\circ\)

\(\theta = 13.4495^\circ \approx 13.4^\circ\)

For the second solution:

\(2\theta + 22.208^\circ = 180^\circ - 49.107^\circ\)

\(2\theta = 180^\circ - 49.107^\circ - 22.208^\circ\)

\(2\theta = 108.685^\circ\)

\(\theta = 54.3425^\circ \approx 54.3^\circ\)

(i) Start with the equation:

\(\sqrt{2} \csc x + \cot x = \sqrt{3}\)

Express \(\csc x\) and \(\cot x\) in terms of sine and cosine:

\(\sqrt{2} \left( \frac{1}{\sin x} \right) + \frac{\cos x}{\sin x} = \sqrt{3}\)

Combine into a single fraction:

\(\frac{\sqrt{2} + \cos x}{\sin x} = \sqrt{3}\)

Multiply through by \(\sin x\):

\(\sqrt{2} + \cos x = \sqrt{3} \sin x\)

Rearrange to:

\(\sqrt{3} \sin x - \cos x = \sqrt{2}\)

Use the identity:

\(a \sin \theta - b \cos \theta \equiv R \sin(\theta - \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = \sqrt{3}\) and \(b = 1\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{1}{\sqrt{3}}\)

\(\alpha = 30^\circ\)

Thus, \(\sqrt{3} \sin x - \cos x = 2 \sin(x - 30^\circ)\).

(ii) Solve \(2 \sin(x - 30^\circ) = \sqrt{2}\):

\(\sin(x - 30^\circ) = \frac{\sqrt{2}}{2}\)

\(x - 30^\circ = 45^\circ\) or \(x - 30^\circ = 135^\circ\)

\(x = 75^\circ\) or \(x = 165^\circ\)

(i) Expand \(2 \sin(x - 30^\circ)\) using the angle subtraction formula:

\(2 \sin(x - 30^\circ) = 2(\sin x \cos 30^\circ - \cos x \sin 30^\circ)\)

\(= \sqrt{3} \sin x - \cos x\)

Combine terms: \(\sqrt{3} \sin x - 2 \cos x\).

Using the identity:

\(a \sin \theta + b \cos \theta \equiv R \sin(\theta + \alpha)\)

\(R = \sqrt{a^2 + b^2} = \sqrt{(\sqrt{3})^2 + (-2)^2} = \sqrt{7}\)

\(\tan \alpha = \frac{b}{a} = \frac{-2}{\sqrt{3}}\)

\(\alpha = \tan^{-1}\left(\frac{-2}{\sqrt{3}}\right) \approx 49.11^\circ\)

(ii) Solve \(\sqrt{3} \sin x - 2 \cos x = 1\):

\(R \sin(x - \alpha) = 1\)

\(\sqrt{7} \sin(x - 49.11^\circ) = 1\)

\(\sin(x - 49.11^\circ) = \frac{1}{\sqrt{7}}\)

\(x - 49.11^\circ = \sin^{-1}\left(\frac{1}{\sqrt{7}}\right) \approx 22.21^\circ\)

\(x = 22.21^\circ + 49.11^\circ = 71.3^\circ\)

(i) To express \(8 \cos \theta - 15 \sin \theta\) in the form \(R \cos(\theta + \alpha)\), we use the identity:

\(a \cos \theta + b \sin \theta \equiv R \cos(\theta - \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 8\) and \(b = -15\).

Calculate \(R\):

\(R = \sqrt{8^2 + (-15)^2} = \sqrt{64 + 225} = \sqrt{289} = 17\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{-15}{8}\)

\(\alpha = \tan^{-1}\left(\frac{-15}{8}\right) \approx 61.93^\circ\)

(ii) Solve \(8 \cos 2x - 15 \sin 2x = 4\).

Using the result from part (i), express the equation as:

\(17 \cos(2x + 61.93^\circ) = 4\)

\(\cos(2x + 61.93^\circ) = \frac{4}{17}\)

\(2x + 61.93^\circ = \cos^{-1}\left(\frac{4}{17}\right) \approx 76.39^\circ\)

\(2x = 76.39^\circ - 61.93^\circ = 14.46^\circ\)

\(x = \frac{14.46^\circ}{2} = 7.23^\circ\)

For the second solution:

\(2x + 61.93^\circ = 360^\circ - 76.39^\circ = 283.61^\circ\)

\(2x = 283.61^\circ - 61.93^\circ = 221.68^\circ\)

\(x = \frac{221.68^\circ}{2} = 110.84^\circ\)

Thus, the solutions are \(x = 7.2^\circ\) and \(x = 110.8^\circ\).

(i) We want to express \(\sqrt{5} \cos x + 2 \sin x\) in the form \(R \cos(x - \alpha)\).

Using the identity:

\(\begin{aligned} a \cos \theta + b \sin \theta &\equiv R \cos(\theta - \alpha), \\ R &= \sqrt{a^2 + b^2}, \\ \tan \alpha &= \frac{b}{a}, \quad 0^\circ < \alpha < 90^\circ. \end{aligned}\)

Here, \(a = \sqrt{5}\) and \(b = 2\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{5})^2 + 2^2} = \sqrt{5 + 4} = \sqrt{9} = 3.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{2}{\sqrt{5}},\)

\(\alpha = \tan^{-1}\left(\frac{2}{\sqrt{5}}\right) \approx 41.81^\circ.\)

(ii) Solve \(\sqrt{5} \cos \frac{1}{2}x + 2 \sin \frac{1}{2}x = 1.2\).

We have \(R \cos\left(\frac{1}{2}x - 41.81^\circ\right) = 1.2\).

\(3 \cos\left(\frac{1}{2}x - 41.81^\circ\right) = 1.2\)

\(\cos\left(\frac{1}{2}x - 41.81^\circ\right) = 0.4\)

\(\frac{1}{2}x - 41.81^\circ = \cos^{-1}(0.4) \approx 66.42^\circ\)

\(\frac{1}{2}x = 66.42^\circ + 41.81^\circ = 108.23^\circ\)

\(x = 2 \times 108.23^\circ = 216.46^\circ\)

Rounding to one decimal place, \(x = 216.5^\circ\).

(i) To express \(3 \sin \theta + 2 \cos \theta\) in the form \(R \sin(\theta + \alpha)\), we use the identity:

\(a \sin \theta + b \cos \theta \equiv R \sin(\theta + \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 3\) and \(b = 2\).

Calculate \(R\):

\(R = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{2}{3}\)

\(\alpha = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.69^\circ\)

(ii) Solve \(3 \sin \theta + 2 \cos \theta = 1\):

Using the expression from part (i):

\(R \sin(\theta + \alpha) = 1\)

\(\sqrt{13} \sin(\theta + 33.69^\circ) = 1\)

\(\sin(\theta + 33.69^\circ) = \frac{1}{\sqrt{13}}\)

\(\theta + 33.69^\circ = \sin^{-1}\left(\frac{1}{\sqrt{13}}\right)\)

\(\sin^{-1}\left(\frac{1}{\sqrt{13}}\right) \approx 16.10^\circ\)

\(\theta + 33.69^\circ = 16.10^\circ\) or \(\theta + 33.69^\circ = 180^\circ - 16.10^\circ\)

\(\theta = 16.10^\circ - 33.69^\circ = -17.59^\circ\) (not in range)

\(\theta = 180^\circ - 16.10^\circ - 33.69^\circ = 130.21^\circ\)

Thus, \(\theta = 130.2^\circ\).

(i) Start with the given equation:

\(\sec \theta + 2 \csc \theta = 3 \csc 2\theta\)

Convert to sines and cosines:

\(\frac{1}{\cos \theta} + \frac{2}{\sin \theta} = \frac{3}{2 \sin \theta \cos \theta}\)

Multiply through by \(2 \sin \theta \cos \theta\):

\(2 \sin \theta + 4 \cos \theta = 3\)

(ii) Express \(2 \sin \theta + 4 \cos \theta\) in the form \(R \sin(\theta + \alpha)\):

\(R = \sqrt{2^2 + 4^2} = \sqrt{20} \approx 4.47\)

\(\tan \alpha = \frac{4}{2} = 2\)

\(\alpha = \tan^{-1}(2) \approx 63.43^\circ\)

(iii) Solve \(\sec \theta + 2 \csc \theta = 3 \csc 2\theta\):

Using the result from (i), solve \(2 \sin \theta + 4 \cos \theta = 3\):

\(\sin(\theta + 63.43^\circ) = \frac{3}{4.47}\)

\(\theta + 63.43^\circ = \sin^{-1}\left(\frac{3}{4.47}\right)\)

\(\theta = 74.4^\circ, 338.7^\circ\)

(i) Expand \(\cos(x + 45^\circ)\):

\(\cos(x + 45^\circ) = \frac{\cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}}\)

Express \(\cos(x + 45^\circ) - \sqrt{2} \sin x\):

\(\frac{\cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}} - \sqrt{2} \sin x = \frac{\cos x}{\sqrt{2}} - \left(\frac{1}{\sqrt{2}} + \sqrt{2}\right) \sin x\)

Let \(a = \frac{1}{\sqrt{2}}\) and \(b = -\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)\). Then:

\(R = \sqrt{a^2 + b^2} = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)\right)^2} = 2.236\)

\(\tan \alpha = \frac{b}{a} = \frac{-\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)}{\frac{1}{\sqrt{2}}}\)

\(\alpha = \tan^{-1}\left(\frac{-\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)}{\frac{1}{\sqrt{2}}}\right) = 71.57^\circ\)

(ii) Solve \(\cos(x + 45^\circ) - \sqrt{2} \sin x = 2\):

Using the expression from (i):

\(R \cos(x + \alpha) = 2\)

\(2.236 \cos(x + 71.57^\circ) = 2\)

\(\cos(x + 71.57^\circ) = \frac{2}{2.236}\)

\(x + 71.57^\circ = \cos^{-1}\left(\frac{2}{2.236}\right)\)

\(x + 71.57^\circ = 26.57^\circ \text{ or } 333.43^\circ\)

\(x = 315^\circ \text{ or } 261.9^\circ\)