Sketching Graphs

Video lesson · from 1:07:471 worked example

One lesson video covers all of Trigonometry, so it opens at 1:07:47 for this topic — not from the beginning.

Theory

Once you have converted $a\sin x + b\cos x$ into a single wave form, such as $k\cos(x - \alpha)$ , sketch it by applying the phase shift $\alpha$ to the $\cos$ or $\sin$ curve multiplied by amplitude $k$ .

⚠️ Common Examiner Traps

This is one of the lowest-scoring questions in the course. Most candidates gain no marks at all on the sketching part. The three points below are exactly what goes wrong.
Translate the right way: shifting the graph in the wrong direction is the most common error. In $k\cos(x - \alpha)$ the graph moves right by $\alpha$ ; a $+\alpha$ moves it left.
Check both ends of the domain: very few candidates consider the height of the graph at each end. Substitute the first and last $x$ values of the stated domain and plot those points — the curve rarely starts or finishes at a maximum.
Mark the key features: maximum $k$ , minimum $-k$ , and the roots. Use the axes provided, and the scale printed on them.

Worked examples

Example 1

Sketch the graph of $y = \sin x^\circ + \sqrt{3}\cos x^\circ$ for $0 \leq x^\circ \leq 360$ .

1. Convert to a single valid form, e.g., $k\cos(x - \alpha)$ .

k\cos(x - \alpha) = k\cos x^\circ\cos\alpha^\circ + k\sin x^\circ\sin\alpha^\circ

\begin{aligned} k\cos\alpha &= \sqrt{3} \\ k\sin\alpha &= 1 \end{aligned}

k = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2

\tan\alpha = \frac{1}{\sqrt{3}} \implies \alpha = 30^\circ \text{ (Q1)}

Equation is $y = 2\cos(x - 30^\circ)$ .

2. Properties of the curve:

Max value: 2 (occurs when $x - 30 = 0 \implies x = 30^\circ$ )
Min value: -2 (occurs when $x - 30 = 180 \implies x = 210^\circ$ )
Roots: $\cos(x - 30) = 0 \implies x - 30 = 90 \implies x = 120^\circ$ and $x - 30 = 270 \implies x = 300^\circ$
y-intercept: let $x = 0$ , $y = 2\cos(-30^\circ) = \sqrt{3} \approx 1.73$