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Let's go through each part of the problem in detail:

### (i) Sketch the curve \( y = \sin x \) for \( 0^\circ \leq x \leq 360^\circ \).

The graph of \( y = \sin x \) within the interval \( 0^\circ \leq x \leq 360^\circ \) will have:
- A starting point at \( (0^\circ, 0) \),
- A peak at \( (90^\circ, 1) \),
- Crosses the x-axis at \( 180^\circ \),
- A trough at \( (270^\circ, -1) \),
- Returns to zero at \( (360^\circ, 0) \).

The graph will look like a wave that oscillates between -1 and 1, peaking and dipping at the points mentioned.

### (ii) Solve the equation \( \sin x = 0.5 \) for \( 0^\circ \leq x \leq 360^\circ \), and illustrate the two roots on your sketch.

For \( \sin x = 0.5 \):
1. The principal angle (one solution within \( 0^\circ \leq 360^\circ \)) for \( \sin^{-1}(0.5) \) is \( 30^\circ \).
2. Since \( \sin \) is positive in the first and second quadrants, the other solution is \( 180^\circ - 30^\circ = 150^\circ \).

Thus, the two roots within \( 0^\circ \leq x \leq 360^\circ \) are:
\[ x = 30^\circ \text{ and } x = 150^\circ. \]

On the sketch, mark these points where the curve \( y = \sin x \) intersects the line \( y = 0.5 \).

### (iii) State the other roots for \( \sin x = 0.5 \), given that \( x \) is no longer restricted to values between \( 0^\circ \) and \( 360^\circ \).

For \( \sin x = 0.5 \), there are infinitely many solutions because the sine function is periodic with a period of \( 360^\circ \). The general solution for \( \sin x = 0.5 \) is:
\[
x = 30^\circ + 360^\circ n \quad \text{and} \quad x = 150^\circ + 360^\circ n
\]
where \( n \) is any integer. These represent all the possible angles that satisfy \( \sin x = 0.5 \).

### (iv) Write down, without using your calculator, the value of \( \sin 330^\circ \).

To find \( \sin 330^\circ \):
- Note that \( 330^\circ \) is in the fourth quadrant.
- In the fourth quadrant, \( \sin \) values are negative.
- \( 330^\circ \) is \( 30^\circ \) less than \( 360^\circ \), so it is equivalent to \( -\sin 30^\circ \).

Since \( \sin 30^\circ = 0.5 \), we have:
\[
\sin 330^\circ = -0.5.
\]

Would you like any further explanations on these parts, or assistance with sketching the graph?

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Here are 5 related questions to extend your understanding:

1. What is the general solution for \( \cos x = 0.5 \)?
2. How would the graph of \( y = \cos x \) differ from \( y = \sin x \) within \( 0^\circ \leq x \leq 360^\circ \)?
3. Why are there only two solutions for \( \sin x = 0.5 \) within \( 0^\circ \leq x \leq 360^\circ \)?
4. What are the roots of \( \sin x = -0.5 \) within \( 0^\circ \leq x \leq 360^\circ \)?
5. How does the amplitude affect the range of a sine or cosine function?

**Tip:** Remember that trigonometric functions are periodic, meaning solutions often repeat at regular intervals.

Solve the trigonometric problem with multiple parts involving the function y = sin x, including sketching, solving equations, finding additional roots, and evaluating sin 330°.

This problem involves sketching y = sin x from 0° to 360°, solving sin x = 0.5 within this range, finding additional roots, and evaluating sin 330°. Ideal for students in Grades 9-11 studying trigonometry.

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