Additional Mathematics 4037/12 — October/November 2019

Walkthrough

The function $y = 2\cos 3x - 1$ is built from the basic cosine wave by three transformations. The factor $2$ in front stretches the wave vertically, so instead of running from $-1$ to $1$ it runs from $-2$ to $2$. The $-1$ at the end slides the whole wave down one unit, so the wave now oscillates between $-3$ and $1$ about the midline $y = -1$. The factor $3$ inside compresses the wave horizontally, so one full cycle takes $360^\circ \div 3 = 120^\circ$ instead of $360^\circ$ — the interval $-90^\circ$ to $90^\circ$ contains one and a half cycles.

To place the curve accurately, evaluate $y$ at the grid angles:

$$y(0) = 2\cos 0 - 1 = 1, \qquad y(\pm 30^\circ) = 2\cos 90^\circ - 1 = -1$$ $$y(\pm 60^\circ) = 2\cos 180^\circ - 1 = -3, \qquad y(\pm 90^\circ) = 2\cos 270^\circ - 1 = -1$$

Since cosine is even, the curve is symmetric about the $y$-axis. Join the points smoothly: the curve starts at $(-90^\circ, -1)$, rises to its peak $(0, 1)$, falls through $(30^\circ, -1)$ to the trough $(60^\circ, -3)$, and rises back to $(90^\circ, -1)$.

Key Takeaways

• For $y = a\cos(bx) + c$: the amplitude is $|a|$, the period is $360^\circ/|b|$, and the midline is $y = c$.
• A horizontal compression by $b$ means more cycles fit in the same interval.
• Evaluating at equally spaced key angles (multiples of a quarter period) gives reliable anchor points for a sketch.

Common Mistakes

• Forgetting the vertical translation and sketching the wave oscillating between $-2$ and $2$ instead of $-3$ and $1$.
• Ignoring the factor of $3$ and drawing a wave with period $360^\circ$, which would not reach the minima at $\pm 60^\circ$.
• Drawing straight line segments between points instead of a smooth curve — the mark scheme requires "attempt at a curve".
• Missing the end points at $(\pm 90^\circ, -1)$; the mark scheme awards a specific mark for starting and finishing there.
• Marking the $y$-intercept incorrectly — it is $(0, 1)$, not $(0, -1)$ or $(0, 2)$.

Things to Be Careful About

• The mark scheme pays one mark for the $y$-intercept $(0, 1)$ (with a graph present), one for the correct end points $(\pm 90^\circ, -1)$, and one for the whole curve passing through $(\pm 30^\circ, -1)$ and $(\pm 60^\circ, -3)$ — all these features must appear.
• The curve must be smooth and symmetric about the $y$-axis; a jagged or asymmetric sketch loses the final mark.
• Work in degrees here, since the axis is marked in degrees.

Walkthrough

The amplitude of a cosine (or sine) wave is half the vertical distance between its maximum and minimum — equivalently, the multiplier of the trig function. In $y = 2\cos 3x - 1$ the multiplier is $2$, so the wave reaches $2$ above and $2$ below its midline $y = -1$, i.e. between $1$ and $-3$. The $-1$ translation and the $3$ inside the cosine do not affect the amplitude.

Key Takeaways

• Amplitude $= |a|$ in $y = a\cos(bx) + c$; translations and horizontal scaling do not change it.

Common Mistakes

• Giving $3$ (the horizontal factor) or $1$ (the vertical shift) instead of $2$.
• Giving the range $(-3$ to $1)$ or half the full height $3$ instead of the amplitude $2$.

Things to Be Careful About

• This is a single B1 mark — the answer must be exactly $2$.

Walkthrough

The factor $3$ inside the cosine compresses the basic $360^\circ$ cycle into one third of the space, so the period is $360^\circ \div 3 = 120^\circ$. This matches the sketch in part (i): the curve completes one full up-and-down pattern every $120^\circ$, so the interval from $-90^\circ$ to $90^\circ$ contains one and a half cycles.

Key Takeaways

• Period $= 360^\circ/|b|$ (or $2\pi/|b|$ in radians) for $y = a\cos(bx) + c$.
• The amplitude and vertical shift do not affect the period.

Common Mistakes

• Multiplying by $3$ instead of dividing, giving $1080^\circ$.
• Giving the amplitude $2$ or the vertical shift $1$ instead of the period.

Things to Be Careful About

• The mark scheme accepts $120^\circ$ or $\frac{2\pi}{3}$ — either exact form scores the B1.