Trigonometry: Question 7

Worked solution

Part (a): Maximum and minimum values of $y$

For any $x$, $\cos(2x)$ takes every value in the interval: $-1 \le \cos(2x) \le 1$

Multiplying by $-4$ reverses this range: $-4 \le -4\cos(2x) \le 4$

Adding the vertical shift of $5$: $5-4 \le y \le 5+4 \quad\Longrightarrow\quad 1 \le y \le 9$

So the maximum value of $y$ is $9$ and the minimum value of $y$ is $1$.

Part (b): Where the maximum occurs

The maximum $y=9$ occurs when $-4\cos(2x)=4$, i.e. when $\cos(2x)=-1$.

Since $0° \le x \le 360°$, the doubled angle satisfies $0° \le 2x \le 720°$. Within this extended range, $\cos(2x)=-1$ at: $2x = 180°, \ 540°$ (the next value, $900°$, lies outside the $720°$ limit).

Dividing by $2$: $x = 90°,\ 270°$

Check: at $x=90°$, $2x=180°$, $\cos 180°=-1$, so $y=5-4(-1)=9$ ✓. At $x=270°$, $2x=540°\equiv180°$, $\cos 540°=-1$, so $y=9$ ✓.

Part (c): Where the minimum occurs

The minimum $y=1$ occurs when $-4\cos(2x)=-4$, i.e. when $\cos(2x)=1$.

Within $0° \le 2x \le 720°$, $\cos(2x)=1$ at: $2x = 0°,\ 360°,\ 720°$ All three lie within the closed interval $0°$ to $720°$ (including both endpoints).

Dividing by $2$: $x = 0°,\ 180°,\ 360°$

Check: at $x=0°$, $\cos 0°=1$, $y=5-4=1$ ✓. At $x=180°$, $2x=360°$, $\cos 360°=1$, $y=1$ ✓. At $x=360°$, $2x=720°$, $\cos 720°=\cos 0°=1$, $y=1$ ✓.

Final answers

• (a) Maximum $y=\boxed{9}$, minimum $y=\boxed{1}$
• (b) Maximum occurs at $x = \boxed{90°,\ 270°}$
• (c) Minimum occurs at $x = \boxed{0°,\ 180°,\ 360°}$