Question: State the amplitude, period, and range of $f(x) = 3\cos\!\left(\dfrac{x}{2}\right) - 1$.

Step 1. Compare with $y = a\cos(bx) + c$:

• $a = 3$
• $b = \dfrac{1}{2}$
• $c = -1$

Step 2. Read off:

• Amplitude $= |a| = 3$
• Period $= \dfrac{2\pi}{|b|} = \dfrac{2\pi}{1/2} = 4\pi$
• Range: oscillates around $c = -1$ with amplitude $3$, so range $= [-1 - 3, -1 + 3] = [-4, 2]$.

Answer: Amplitude $3$, period $4\pi$, range $[-4, 2]$.

Tip: Period is "how long for one full cycle" $= \dfrac{2\pi}{|b|}$ in radians, or $\dfrac{360°}{|b|}$ in degrees.

Question: The curve $y = a\sin(bx) + c$ has a minimum at $\left(\dfrac{\pi}{6}, -3\right)$ and a maximum at $\left(\dfrac{\pi}{2}, 7\right)$. Find $a$, $b$, $c$.

Step 1. Amplitude $a$ from max/min:

$$a = \dfrac{\text{max} - \text{min}}{2} = \dfrac{7 - (-3)}{2} = 5$$

Step 2. Vertical shift $c$:

$$c = \dfrac{\text{max} + \text{min}}{2} = \dfrac{7 + (-3)}{2} = 2$$

Step 3. Period $b$ from distance between consecutive max and min: that's half a period.

Distance $= \dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{\pi}{3}$, so half-period $= \dfrac{\pi}{3}$ → period $= \dfrac{2\pi}{3}$.

$$b = \dfrac{2\pi}{\text{period}} = \dfrac{2\pi}{2\pi/3} = 3$$

Answer: $a = 5$, $b = 3$, $c = 2$.

Tip: "Min came first, then max" → sine starts BELOW the centre line, so $a > 0$ (positive sine that's been shifted to start at min). If max came first instead, $a$ might be negative.

Question: Sketch $y = 1 - 3\sin 4x$ for $0° \leq x \leq 180°$, marking key points.

Step 1. Identify features:

• Amplitude $= |{-3}| = 3$, vertical shift $= 1$
• Period $= \dfrac{360°}{4} = 90°$
• Range $= [1 - 3, 1 + 3] = [-2, 4]$
• The negative coefficient flips the sine: starts at $(0, 1)$ but FALLS first (rather than rises).

Step 2. Key points within one period ($0° \to 90°$):

• $x = 0°$: $y = 1$ (centre)
• $x = 22.5°$: $\sin 4x = 1$ → $y = 1 - 3 = -2$ (min)
• $x = 45°$: $\sin 4x = 0$ → $y = 1$ (centre)
• $x = 67.5°$: $\sin 4x = -1$ → $y = 1 + 3 = 4$ (max)
• $x = 90°$: $\sin 4x = 0$ → $y = 1$ (centre)

Step 3. Repeat the pattern for $90° \leq x \leq 180°$: same shape, second full cycle.

Tip: Mark $x$-values where $4x$ is a multiple of $90°$ — those are the crests, troughs, and zero-crossings. Two full periods fit in $[0°, 180°]$.

Question: On the same diagram, sketch $y = 4\cos x$ and $y = 2 + 3\sin x$ for $0 \leq x \leq 2\pi$.

$y = 4\cos x$:

• Amplitude $4$, period $2\pi$, no vertical shift.
• $x = 0$: $y = 4$ (max)
• $x = \pi/2$: $y = 0$
• $x = \pi$: $y = -4$ (min)
• $x = 3\pi/2$: $y = 0$
• $x = 2\pi$: $y = 4$ (max)

$y = 2 + 3\sin x$:

• Amplitude $3$, vertical shift $2$, period $2\pi$. Range $[-1, 5]$.
• $x = 0$: $y = 2$
• $x = \pi/2$: $y = 5$ (max)
• $x = \pi$: $y = 2$
• $x = 3\pi/2$: $y = -1$ (min)
• $x = 2\pi$: $y = 2$

Step 3. Both curves drawn on same axes. They intersect at points where $4\cos x = 2 + 3\sin x$ — a useful follow-up question.

Tip: Plot key points (max, min, intercepts) for BOTH curves first, then connect smoothly. Different colours help.

Question: By drawing $y = \dfrac{2}{\pi}x + 1$, find the number of solutions of $\pi\sin 3x - x = 0$ for $0 \leq x \leq \pi$.

Step 1. Rearrange the equation:

$$\pi\sin 3x = x \;\Rightarrow\; \sin 3x = \dfrac{x}{\pi}$$

To use the suggested line, rewrite as $1 - \sin 3x = 1 - \dfrac{x}{\pi}$... hmm, let me reconsider.

Actually rewrite $\pi\sin 3x = x$ as $\sin 3x = \dfrac{x}{\pi}$. Sketch $y = \sin 3x$ and $y = \dfrac{x}{\pi}$ on the same axes; their intersections are solutions.

But the auxiliary line specified is $y = \dfrac{2}{\pi}x + 1$. Restate: rearrange the original equation to have one side match this line. Try $\sin 3x + \dfrac{x}{\pi} = \dfrac{2}{\pi}x$ ... actually the cleanest version: rewrite as $1 + \sin 3x = 1 + \dfrac{x}{\pi}$, then both sides involve known shapes.

Step 2. Sketch $y = \sin 3x$ for $0 \leq x \leq \pi$: three full periods squashed into $[0, \pi]$ (since period $= \dfrac{2\pi}{3}$, you get $1.5$ cycles). The line $y = \dfrac{x}{\pi}$ goes from $(0, 0)$ to $(\pi, 1)$.

Step 3. Count intersections — typically the line crosses the sine curve at 3 points across the interval.

Tip: Auxiliary-line problems work by reshaping the equation into "trig curve = straight line", then counting intersection points.

Question: Show that $6\sin^2 x - 4\cos^2 x \equiv 1 - 5\cos 2x$. Hence state the amplitude and period of $f(x) = 6\sin^2 x - 4\cos^2 x$.

Step 1. Use $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ and $\cos^2 x = \dfrac{1 + \cos 2x}{2}$:

$$6\sin^2 x - 4\cos^2 x = 6 \cdot \dfrac{1 - \cos 2x}{2} - 4 \cdot \dfrac{1 + \cos 2x}{2}$$ $$= 3(1 - \cos 2x) - 2(1 + \cos 2x)$$ $$= 3 - 3\cos 2x - 2 - 2\cos 2x$$ $$= 1 - 5\cos 2x \;\checkmark$$

Step 2. Read off the sinusoidal form:

• Amplitude $= |-5| = 5$
• Period $= \dfrac{2\pi}{2} = \pi$
• Vertical shift $= 1$ → range $[1 - 5, 1 + 5] = [-4, 6]$

Why this works: $\sin^2 x$ and $\cos^2 x$ are NOT sinusoids — they don't have a clean amplitude or period. The double-angle identity converts them into a standard $a + b\cos 2x$ form, which IS sinusoidal.

Question: The height $h$ (metres) of a wind turbine blade tip is modelled by $h = 80 - 50\cos\!\left(\dfrac{\pi t}{2}\right)$, where $t$ is in minutes. Find the duration during one minute that the blade tip is above $40$ m.

Step 1. Set up the inequality:

$$80 - 50\cos\!\left(\dfrac{\pi t}{2}\right) > 40$$ $$-50\cos\!\left(\dfrac{\pi t}{2}\right) > -40$$ $$\cos\!\left(\dfrac{\pi t}{2}\right) < 0.8$$

Step 2. $\cos\theta = 0.8 \Rightarrow \theta = \cos^{-1}(0.8) \approx 0.6435$ rad.

Cosine is BELOW $0.8$ when $\theta \in (0.6435, 2\pi - 0.6435)$, i.e. $\theta \in (0.6435, 5.640)$.

Step 3. Convert to $t$: $\dfrac{\pi t}{2} \in (0.6435, 5.640)$:

$$t \in \left(\dfrac{2 \times 0.6435}{\pi}, \dfrac{2 \times 5.640}{\pi}\right) \approx (0.410, 3.59)$$

So in one full period (which is $\dfrac{2\pi}{\pi/2} = 4$ minutes), the blade is above $40$ m from $t \approx 0.41$ to $t \approx 3.59$, a duration of $3.59 - 0.41 \approx 3.18$ minutes.

Step 4. In ONE minute ($t \in [0, 1]$), the blade is above $40$ m from $t = 0.41$ to $t = 1$, a duration of $\approx 0.59$ minutes (35 seconds).

Tip: Always solve the equation (boundary) first, then use the cosine/sine shape to identify whether you want the "inside" or "outside" of the boundary interval.