Tasks 04-04 - Trigonometric Functions

Section 04: Advanced Functions

A: Skills

Problem 1: Angle Conversion (x)

Convert 225° to radians.
Convert $\frac{7\pi}{6}$ radians to degrees.
Convert −120° to radians.
Convert $\frac{3\pi}{4}$ radians to degrees.
Compute the length of the arc that the angle $\frac{3\pi}{4}$ cuts out of a circle with a radius of 8 cm, exactly and rounded to two decimal places.

Solution

Conversion: $180° = \pi$ radians, so multiply by $\frac{\pi}{180°}$ to get radians and by $\frac{180°}{\pi}$ to get degrees.

$225° \cdot \frac{\pi}{180°} = \frac{225\pi}{180} = \frac{5\pi}{4}$
$\frac{7\pi}{6} \cdot \frac{180°}{\pi} = \frac{7 \cdot 180°}{6} = 210°$
$-120° \cdot \frac{\pi}{180°} = -\frac{120\pi}{180} = -\frac{2\pi}{3}$
$\frac{3\pi}{4} \cdot \frac{180°}{\pi} = \frac{3 \cdot 180°}{4} = 135°$
With the angle in radians, the arc length is $s = r\theta$:

\[s = 8 \cdot \frac{3\pi}{4} = 6\pi \approx 18.85 \text{ cm}\]

Check: $135°$ is $\frac{135}{360} = \frac{3}{8}$ of a full turn, and $\frac{3}{8}$ of the circumference $2\pi \cdot 8 = 16\pi$ is $6\pi$ ✓

Problem 2: Exact Trigonometric Values (x)

Find the exact values without a calculator.

$\cos\left(\frac{\pi}{6}\right)$
$\sin\left(\frac{\pi}{4}\right)$
$\tan\left(\frac{\pi}{3}\right)$
$\cos(3\pi)$
$\sin\left(\frac{5\pi}{2}\right)$

Hint: $\sin(30°) = \frac{1}{2}$ and $\sin(60°) = \frac{\sqrt{3}}{2}$. The point $(\cos(\theta), \sin(\theta))$ lies on the unit circle $x^2 + y^2 = 1$, so $\cos^2(\theta) + \sin^2(\theta) = 1$.

Solution

The point for the angle $\theta$ on the unit circle is $(\cos(\theta), \sin(\theta))$.

$\frac{\pi}{6} = 30°$. From $\sin(30°) = \frac{1}{2}$ and $\cos^2(\theta) + \sin^2(\theta) = 1$: $\cos^2(30°) = 1 - \frac{1}{4} = \frac{3}{4}$. The point lies to the right of the $y$-axis, so the cosine is positive:

\[\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\]

$\frac{\pi}{4} = 45°$ lies exactly halfway between the axes, so the point has equal coordinates $(a, a)$ with $a^2 + a^2 = 1$, i.e. $a^2 = \frac{1}{2}$:

\[\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\]

$\frac{\pi}{3} = 60°$. From $\sin(60°) = \frac{\sqrt{3}}{2}$: $\cos^2(60°) = 1 - \frac{3}{4} = \frac{1}{4}$. The point lies to the right of the $y$-axis, so $\cos(60°) = \frac{1}{2}$ and the point is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$, so

\[\tan\left(\frac{\pi}{3}\right) = \frac{\sin\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right)} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}\]

$3\pi = \pi + 2\pi$. A full turn ($2\pi$) leads back to the same point, so $3\pi$ gives the same point as $\pi$, the point $(-1, 0)$ on the left of the circle:

\[\cos(3\pi) = \cos(\pi) = -1\]

$\frac{5\pi}{2} = \frac{\pi}{2} + 2\pi$, the point $(0, 1)$ at the top of the circle:

\[\sin\left(\frac{5\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1\]

Problem 3: Unit Circle Coordinates (xx)

Find the exact coordinates $(\cos(\theta), \sin(\theta))$ of the point on the unit circle for each angle. Check one of your points with $\cos^2(\theta) + \sin^2(\theta) = 1$.

$\theta = \frac{2\pi}{3}$
$\theta = \frac{5\pi}{4}$
$\theta = \frac{11\pi}{6}$
$\theta = \frac{4\pi}{3}$

Hint: the unit circle is symmetric. Mirroring the point for $\theta$ at the $y$-axis gives the point for $\pi - \theta$: $(-\cos(\theta), \sin(\theta))$. Mirroring it at the $x$-axis gives the point for $-\theta$: $(\cos(\theta), -\sin(\theta))$. Half a turn leads to the opposite point, the point for $\theta + \pi$: $(-\cos(\theta), -\sin(\theta))$. A full turn ($2\pi$) does not change the point.

Solution

Write each angle with one of the angles $\frac{\pi}{6}$, $\frac{\pi}{4}$ or $\frac{\pi}{3}$ and use the hint.

$\frac{2\pi}{3} = \pi - \frac{\pi}{3}$: mirror the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ for $\frac{\pi}{3}$ at the $y$-axis.

\[\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\]

$\frac{5\pi}{4} = \frac{\pi}{4} + \pi$: the opposite point of $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.

\[\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)\]

$\frac{11\pi}{6} = -\frac{\pi}{6} + 2\pi$: the same point as for $-\frac{\pi}{6}$, i.e. the point $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ for $\frac{\pi}{6}$ mirrored at the $x$-axis.

\[\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)\]

$\frac{4\pi}{3} = \frac{\pi}{3} + \pi$: the opposite point of $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.

\[\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)\]

Check for b): $\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} + \frac{2}{4} = 1$ ✓

Problem 4: Analyzing Trigonometric Functions (xx)

Determine the amplitude, the period, the midline and the range of each function.

$y = 5\cos(x) - 2$
$y = \sin\left(\frac{x}{3}\right)$
$y = -2\cos(\pi x)$
$y = 1.5\sin\left(\frac{\pi}{4}x\right) + 4$

Solution

For $y = A\sin(Bx) + D$ or $y = A\cos(Bx) + D$: amplitude $|A|$, period $\frac{2\pi}{B}$, midline $y = D$, range $[D - |A|; D + |A|]$.

Amplitude 5, period $2\pi$, midline $y = -2$, range $[-2 - 5; -2 + 5] = [-7; 3]$
$B = \frac{1}{3}$: amplitude 1, period $\frac{2\pi}{1/3} = 6\pi$, midline $y = 0$, range $[-1; 1]$
Amplitude $|-2| = 2$ (the minus sign reflects the graph at the $x$-axis, the amplitude stays positive), period $\frac{2\pi}{\pi} = 2$, midline $y = 0$, range $[-2; 2]$
Amplitude 1.5, period $\frac{2\pi}{\pi/4} = 8$, midline $y = 4$, range $[2.5; 5.5]$

Problem 5: Graphing Transformations (xx)

to d): Name the transformations of $y = \sin(x)$ or $y = \cos(x)$ in the correct order, give the five key points of one period (start, maximum or minimum, midline, …) and sketch one period.
$y = \cos\left(x - \frac{\pi}{3}\right)$
$y = -2\sin(x) + 1$
$y = \cos\left(\frac{x}{2}\right)$
$y = 2\sin\left(2\left(x - \frac{\pi}{4}\right)\right)$
Show that $-\cos\left(x + \frac{\pi}{2}\right) = \sin(x)$ for every $x$.

Hint for e): $\cos(x) = \sin\left(x + \frac{\pi}{2}\right)$ and $\sin(x + \pi) = -\sin(x)$ for every $x$.

Solution

Order (as for every function): horizontal stretch or compression, horizontal shift, vertical stretch or compression (with reflection), vertical shift.

a) Horizontal shift by $\frac{\pi}{3}$ to the right (minus shifts right). Amplitude 1, period $2\pi$, midline $y = 0$. The key points of $\cos(x)$ move $\frac{\pi}{3}$ to the right:

$x$	$\frac{\pi}{3}$	$\frac{5\pi}{6}$	$\frac{4\pi}{3}$	$\frac{11\pi}{6}$	$\frac{7\pi}{3}$
$y$	1 (max)	0	$-1$ (min)	0	1 (max)

b) Vertical stretch by 2 and reflection at the $x$-axis (factor $-2$), then vertical shift up by 1. Period $2\pi$, midline $y = 1$, range $[-1; 3]$.

$x$	$0$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$	$2\pi$
$y$	1	$-1$ (min)	1	3 (max)	1

The graph starts on the midline and goes down first, because of the reflection.

c) Horizontal stretch by the factor 2 ($B = \frac{1}{2}$), period $\frac{2\pi}{1/2} = 4\pi$. Amplitude 1, midline $y = 0$.

$x$	$0$	$\pi$	$2\pi$	$3\pi$	$4\pi$
$y$	1 (max)	0	$-1$ (min)	0	1 (max)

d) First the horizontal compression by the factor 2: $\sin(2x)$, period $\frac{2\pi}{2} = \pi$. Then the horizontal shift by $\frac{\pi}{4}$ to the right: $\sin\left(2\left(x - \frac{\pi}{4}\right)\right)$. Then the vertical stretch by 2. The period starts at $x = \frac{\pi}{4}$ and ends at $\frac{\pi}{4} + \pi = \frac{5\pi}{4}$:

$x$	$\frac{\pi}{4}$	$\frac{\pi}{2}$	$\frac{3\pi}{4}$	$\pi$	$\frac{5\pi}{4}$
$y$	0	2 (max)	0	$-2$ (min)	0

Check: $y\left(\frac{\pi}{2}\right) = 2\sin\left(2 \cdot \frac{\pi}{4}\right) = 2\sin\left(\frac{\pi}{2}\right) = 2$ ✓

Typical mistake: shifting $\sin(2x)$ by $\frac{\pi}{2}$ instead of $\frac{\pi}{4}$. In $\sin\left(2\left(x - \frac{\pi}{4}\right)\right)$ the shift is the number subtracted from $x$ inside the bracket, after the $B$ is factored out.

e) Replace $x$ by $x + \frac{\pi}{2}$ in the first identity of the hint:

\[\cos\left(x + \frac{\pi}{2}\right) = \sin\left(x + \frac{\pi}{2} + \frac{\pi}{2}\right) = \sin(x + \pi) = -\sin(x)\]

So $-\cos\left(x + \frac{\pi}{2}\right) = -(-\sin(x)) = \sin(x)$ ✓. The cosine shifted $\frac{\pi}{2}$ to the left and reflected at the $x$-axis is exactly the sine curve.

B: Set Up & Check

Problem 6: From Text to Functions (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works.

“A bike rental at a lake rents out the most bikes in July, 380 bikes per day, and the fewest in January, 40 bikes per day. Over the year, the number of bikes per day follows a sinusoidal pattern.” (A sinusoidal pattern is described by a function $A\sin(B(t - C)) + D$ or $A\cos(B(t - C)) + D$, and its graph is called a sinusoid.) Give the amplitude and the midline.

Check: Compute midline + amplitude and midline − amplitude. Do you get the numbers for July and January? Are July and January half a period apart?
“A bouldering hall (a climbing hall with low walls and soft mats instead of ropes) has $V(t) = 300 + 120\sin(Bt)$ visitors on day $t$. The pattern repeats every week.” Give $B$.

Check: Day 1 and day 8 are exactly one week apart. Compute $V(1)$ and $V(8)$ with your $B$ (the indicator R must show). Are they equal?
“The number of players who are online in an online game follows a 24-hour cycle. The cycle starts at its lowest point: at 5:00, 40 thousand players are online. At 17:00, the number is highest with 200 thousand players.” Give the number of players $N(t)$ in thousands, with $t$ in hours after 5:00.

Check: Compute $N(0)$ and $N(12)$. Do you get the lowest and the highest number?
An FSP task says: “Give a function $f_1(t)$ which is the mirror image of $f(t) = 40 + 15\cos\left(\frac{\pi}{4}t\right)$ at its midline and give a reason for your choice.” Give $f_1(t)$ and the reason.

Check: Compute $f(0)$ and $f_1(0)$. Is their average the midline? A classmate answers with $f(-t)$, the mirror image at the $y$-axis. Is $f(-t)$ a different function from $f(t)$?
“A noodle bar’s delivery service receives more than 40 orders per hour only between 18:00 and 22:00.” The number of orders per hour is modelled by a sinusoidal function $N(t)$ with a period of 24 hours, with $t$ in hours after midnight and $0 \leq t < 24$.

Check: Which value must $N$ have at 18:00 and at 22:00? At which time does the delivery service receive the most orders?

Solution

Amplitude $A = \frac{380 - 40}{2} = 170$ bikes, midline $y = \frac{380 + 40}{2} = 210$ bikes per day.

Check: yes. $210 + 170 = 380$ (July) and $210 - 170 = 40$ (January). The pattern repeats every 12 months, and July is 6 months after January: half a period, as it must be between a maximum and a minimum.

Typical mistake: amplitude $380 - 40 = 340$. The amplitude is the distance from the midline to the maximum, so it is only half the difference between the maximum and the minimum. With 340, midline + amplitude would give 550 bikes.
Period 7 days, so

\[B = \frac{2\pi}{7}, \qquad V(t) = 300 + 120\sin\left(\frac{2\pi}{7}t\right)\]

Check: yes. $V(8) = 300 + 120\sin\left(\frac{16\pi}{7}\right) = 300 + 120\sin\left(\frac{2\pi}{7} + 2\pi\right) = V(1) \approx 393.82$. One week adds exactly $2\pi$ to the angle.

Typical mistake: $B = 7$. Then the period is $\frac{2\pi}{7} \approx 0.9$ days, not 7 days, and $V(1) = 300 + 120\sin(7) \approx 378.84$ but $V(8) = 300 + 120\sin(56) \approx 237.41$. The period is $\frac{2\pi}{B}$, so $B = \frac{2\pi}{\text{period}}$.
Amplitude $\frac{200 - 40}{2} = 80$, midline $y = \frac{200 + 40}{2} = 120$, period 24 hours, so $B = \frac{2\pi}{24} = \frac{\pi}{12}$. The cycle starts at its lowest point, so use a negative cosine:

\[N(t) = 120 - 80\cos\left(\frac{\pi}{12}t\right)\]

Check: yes. $N(0) = 120 - 80 \cdot 1 = 40$ and $N(12) = 120 - 80\cos(\pi) = 120 + 80 = 200$.

Typical mistake: $120 - 80\sin\left(\frac{\pi}{12}t\right)$ gives $N(0) = 120$: a sine starts on the midline, not at the lowest point. And $120 + 80\cos\left(\frac{\pi}{12}t\right)$ starts at the highest point.
$f_1(t) = 40 - 15\cos\left(\frac{\pi}{4}t\right)$

Reason: mirroring at the midline $y = 40$ keeps every $t$ and turns a value $40 + d$ into $40 - d$. So the midline stays and the term $15\cos\left(\frac{\pi}{4}t\right)$ changes its sign. In short, $f_1(t) = 80 - f(t)$.

Check: $f(0) = 55$ and $f_1(0) = 25$, with the average $\frac{55 + 25}{2} = 40$ ✓. The classmate’s answer is wrong: $f(-t) = 40 + 15\cos\left(-\frac{\pi}{4}t\right) = f(t)$, because the points for $u$ and $-u$ on the unit circle are mirror images at the $x$-axis and have the same $x$-coordinate, $\cos(-u) = \cos(u)$. The graph of $f$ is symmetric about the $y$-axis, so mirroring it there does not change it.

Typical mistake: $-f(t) = -40 - 15\cos\left(\frac{\pi}{4}t\right)$. This is the mirror image at the $t$-axis, with the midline $y = -40$.
$N(t) > 40$ for $18 < t < 22$, and $N(t) \leq 40$ for all other $t$ with $0 \leq t < 24$.

Check: $N(18) = N(22) = 40$. Just before 18:00, $N$ is at most 40, just after it is above 40, so the graph crosses the line $y = 40$ there. The part of the graph above 40 is the hump around the maximum, and the graph of a sinusoid is symmetric about its maximum. So the most orders come in the middle, at 20:00.

Typical mistake: $18 < N(t) < 22$. The numbers 18 and 22 are times, so they belong to $t$, and 40 is a number of orders, so it belongs to $N(t)$.

Problem 7: One Paragraph, All Conditions (xx)

A dealer sells wood pellets (small pressed sticks of wood that are burnt in special heating systems). Its monthly sales are highest in January with 840 tonnes and lowest in July with 240 tonnes, and the pattern repeats every year. The dealer models the sales $S(t)$ in tonnes per month by a sinusoidal function, with $t$ in months after 1 January ($t = 0$ stands for January). It sells the pellets for 350 EUR per tonne. In every month with sales of more than 690 tonnes, it rents an additional lorry. The dealer wants to know its revenue in each month and the months in which it needs the additional lorry.

List every condition and every function in this paragraph as a mathematical statement. Do not solve anything.

Check: An assistant writes $S(t) = 540 + 300\sin\left(\frac{\pi}{6}t\right)$. Compute $S(0)$ with this term and name the mistake.

Solution

Highest sales in January: $S(0) = 840$
Lowest sales in July: $S(6) = 240$
Period 12 months: $B = \frac{2\pi}{12} = \frac{\pi}{6}$
Amplitude $\frac{840 - 240}{2} = 300$, midline $y = \frac{840 + 240}{2} = 540$
Sales in tonnes per month, starting at the maximum, so with a cosine: $S(t) = 540 + 300\cos\left(\frac{\pi}{6}t\right)$ with $0 \leq t < 12$
Revenue in EUR per month: $R(t) = 350 \cdot S(t)$
Additional lorry: $S(t) > 690$

Check: $S(0) = 540 + 300\sin(0) = 540$, the midline instead of the maximum 840. A sine starts on the midline. For a maximum at $t = 0$, the model needs a cosine.

Typical mistake: $B = 12$. This would make the period $\frac{2\pi}{12} \approx 0.52$ months. The period is 12, so $B = \frac{2\pi}{12}$.

C: Applied

Problem 8: Rotor Blades of a Wind Turbine (xx)

A wind farm operator runs a wind turbine whose hub (the centre of the rotor) is 110 m above the ground. Each rotor blade is 70 m long, and the rotor turns once every 6 seconds at a constant speed. At $t = 0$, blade A points straight down, so its tip is at its lowest point. $h(t)$ is the height of the tip of blade A above the ground in m, with $t$ in seconds. Round heights and speeds to two decimal places.

Determine the midline, the amplitude, the period and $B$. Give the function $h(t)$ and give a reason why a negative cosine fits.

(for verification: $h(t) = 110 - 70\cos\left(\frac{\pi}{3}t\right)$)

Compute the height of the tip after 1, 1.5, 3 and 5 seconds without a calculator. Explain why the tip has the same height after 1 second and after 5 seconds.
State the highest and the lowest height of the tip. Determine all times in the first 20 seconds at which the tip is at its highest point.
A nature protection authority asks for how many seconds per rotation the tip of blade A is lower than 75 m above the ground. Determine this time.

Hint: in $[0; 2\pi]$, $\cos(u) = \frac{1}{2}$ exactly for $u = \frac{\pi}{3}$ and $u = \frac{5\pi}{3}$.

Compute the distance the tip travels in one rotation and its speed in m/s and in km/h (1 m/s = 3.6 km/h). In strong wind, the rotor turns once every 5 seconds. Give the new function $h_2(t)$, name the transformation of $h$ and compute the new speed of the tip in km/h.

Solution

a) The tip moves on a circle around the hub:

Midline: the hub height, $y = 110$ m
Amplitude: the blade length, 70 m (lowest point $110 - 70 = 40$ m, highest point $110 + 70 = 180$ m)
Period: 6 seconds, so $B = \frac{2\pi}{6} = \frac{\pi}{3}$

At $t = 0$ the tip is at its lowest point. A cosine starts at its maximum, a sine on the midline. A negative cosine starts at its minimum:

\[h(t) = 110 - 70\cos\left(\frac{\pi}{3}t\right)\]

Check: $h(0) = 110 - 70 = 40$ ✓ (lowest point)

$h(1) = 110 - 70\cos\left(\frac{\pi}{3}\right) = 110 - 70 \cdot \frac{1}{2} = 75$ m
$h(1.5) = 110 - 70\cos\left(\frac{\pi}{2}\right) = 110 - 0 = 110$ m (level with the hub, blade A is horizontal)
$h(3) = 110 - 70\cos(\pi) = 110 + 70 = 180$ m (highest point, blade A points straight up)
$h(5) = 110 - 70\cos\left(\frac{5\pi}{3}\right) = 110 - 70 \cdot \frac{1}{2} = 75$ m

After 1 second, the tip has left the lowest point 1 second ago. After 5 seconds, it will reach the next lowest point ($t = 6$) 1 second later. The graph is symmetric about each minimum, so both times give the same height: one on the way up, one on the way down.

c) Highest height $110 + 70 = 180$ m, lowest height $110 - 70 = 40$ m.

The tip is at the top when $\cos\left(\frac{\pi}{3}t\right) = -1$, i.e. when $\frac{\pi}{3}t = \pi, 3\pi, 5\pi, \ldots$ This is half a period after each lowest point: $t = 3$, then every 6 seconds. In the first 20 seconds: $t = 3$, $t = 9$ and $t = 15$ seconds.

d) Condition for one rotation, $0 \leq t \leq 6$:

\[110 - 70\cos\left(\frac{\pi}{3}t\right) < 75 \iff -70\cos\left(\frac{\pi}{3}t\right) < -35 \iff \cos\left(\frac{\pi}{3}t\right) > \frac{1}{2}\]

Dividing by the negative number $-70$ reverses the inequality sign.

With $u = \frac{\pi}{3}t$, $t \in [0; 6]$ gives $u \in [0; 2\pi]$. By the hint, $\cos(u) = \frac{1}{2}$ at $u = \frac{\pi}{3}$ and $u = \frac{5\pi}{3}$. Between them, the cosine passes its minimum $-1$ at $u = \pi$, so there it is smaller than $\frac{1}{2}$. The cosine is greater than $\frac{1}{2}$ only near $u = 0$ and $u = 2\pi$:

\[0 \leq u < \frac{\pi}{3} \text{ or } \frac{5\pi}{3} < u \leq 2\pi \iff 0 \leq t < 1 \text{ or } 5 < t \leq 6\]

The tip is lower than 75 m for $1 + 1 = $ 2 seconds per rotation, a third of the time. This matches b): $h(1) = h(5) = 75$ m.

e) In one rotation, the tip travels the circumference of a circle with the radius 70 m (arc length $s = r\theta$ with $\theta = 2\pi$):

\[s = 70 \cdot 2\pi = 140\pi \approx 439.82 \text{ m}\]

Speed: $\frac{140\pi}{6} \approx 73.30$ m/s $= 73.30 \cdot 3.6 \approx 263.89$ km/h.

New period 5 seconds, so $B = \frac{2\pi}{5}$:

\[h_2(t) = 110 - 70\cos\left(\frac{2\pi}{5}t\right)\]

$\frac{2\pi}{5} = 1.2 \cdot \frac{\pi}{3}$, so $h_2(t) = h(1.2t)$: a horizontal compression by the factor 1.2. Midline and amplitude stay the same. New speed: $\frac{140\pi}{5} \approx 87.96$ m/s $\approx 316.67$ km/h.

Problem 9: Water Consumption of a Town (xx)

A town’s waterworks models the water consumption of the town by

\[W(h) = 750 + 250\sin\left(\frac{\pi}{12}(h - 10)\right)\]

in m³ per hour, where $h$ is the time in hours after midnight ($0 \leq h \leq 24$). Round volumes to two decimal places.

State the amplitude, the midline, the period and the phase shift of $W$. Explain what the period and the midline mean for the waterworks.
Determine the times of the highest and the lowest consumption and the consumption at these times.
Compute the consumption at midnight, at 12:00 and at 20:00 without a calculator. Explain why the values at 12:00 and at 20:00 are equal.
The pumps of the waterworks deliver a constant 875 m³ per hour. When the town uses more, a water tower supplies the rest. Determine the time interval in which the water tower is needed, and the largest amount per hour that it must supply.

Hint: in $[0; 2\pi]$, $\sin(u) = \frac{1}{2}$ exactly for $u = \frac{\pi}{6}$ and $u = \frac{5\pi}{6}$.

On hot summer days, the town uses 20 % more water at every hour. Give the function $W_s(h)$ for a hot day with its amplitude and midline. The pumps can deliver at most 1,150 m³ per hour. Decide whether the pumps alone can cover the highest consumption of a hot day.

Solution

a) Amplitude 250 m³ per hour, midline $y = 750$, period $\frac{2\pi}{\pi/12} = 24$ hours, phase shift 10 hours to the right.

Period 24 hours: the consumption follows the same pattern every day.
Midline 750: the consumption swings around 750 m³ per hour. It lies 250 m³ above this value at its highest and 250 m³ below at its lowest.

b) The sine reaches its maximum 1 at $u = \frac{\pi}{2}$ and its minimum $-1$ at $u = -\frac{\pi}{2}$ (or $\frac{3\pi}{2}$):

Highest: $\frac{\pi}{12}(h - 10) = \frac{\pi}{2} \iff h - 10 = 6 \iff h = 16$. At 16:00, $W(16) = 750 + 250 = 1{,}000$ m³ per hour.
Lowest: $\frac{\pi}{12}(h - 10) = -\frac{\pi}{2} \iff h - 10 = -6 \iff h = 4$. At 4:00, $W(4) = 750 - 250 = 500$ m³ per hour.

The minimum comes 12 hours (half a period) before the maximum.

Midnight: $W(0) = 750 + 250\sin\left(-\frac{5\pi}{6}\right) = 750 + 250 \cdot \left(-\frac{1}{2}\right) = 625$ m³ per hour
12:00: $W(12) = 750 + 250\sin\left(\frac{\pi}{6}\right) = 750 + 125 = 875$ m³ per hour
20:00: $W(20) = 750 + 250\sin\left(\frac{5\pi}{6}\right) = 750 + 125 = 875$ m³ per hour

$\sin\left(-\frac{5\pi}{6}\right) = -\sin\left(\frac{5\pi}{6}\right) = -\frac{1}{2}$ (mirror at the $x$-axis), and $\frac{5\pi}{6} = \pi - \frac{\pi}{6}$ has the same sine as $\frac{\pi}{6}$ (mirror at the $y$-axis).

12:00 and 20:00 both lie 4 hours away from the maximum at 16:00. The graph is symmetric about its maximum, so the consumption is the same.

d) Condition:

\[750 + 250\sin\left(\frac{\pi}{12}(h - 10)\right) > 875 \iff \sin\left(\frac{\pi}{12}(h - 10)\right) > \frac{1}{2}\]

With $u = \frac{\pi}{12}(h - 10)$, $h \in [0; 24]$ gives $u \in \left[-\frac{5\pi}{6}; \frac{7\pi}{6}\right]$. For $u < 0$ the sine is negative. For $0 \leq u \leq \frac{7\pi}{6}$, the hint gives the boundaries $\frac{\pi}{6}$ and $\frac{5\pi}{6}$, and between them the sine passes its maximum 1 at $\frac{\pi}{2}$:

\[\frac{\pi}{6} < \frac{\pi}{12}(h - 10) < \frac{5\pi}{6} \iff 2 < h - 10 < 10 \iff 12 < h < 20\]

The water tower is needed from 12:00 to 20:00, i.e. for 8 hours a day. This matches c): at 12:00 and 20:00, the consumption is exactly 875 m³ per hour. The largest amount is needed at 16:00: $1{,}000 - 875 = 125$ m³ per hour.

e) Every value is multiplied by 1.2, a vertical stretch:

\[W_s(h) = 1.2 \cdot W(h) = 900 + 300\sin\left(\frac{\pi}{12}(h - 10)\right)\]

Amplitude 300, midline $y = 900$: both are 20 % larger. The highest consumption is $900 + 300 = 1{,}200$ m³ per hour at 16:00. $1{,}200 > 1{,}150$, so the pumps cannot cover it alone. At 16:00, 50 m³ per hour are missing, which the water tower must supply.

Typical mistake: $W(h) + 150$ (only the midline rises by 20 % of 750). “20 % more at every hour” multiplies every value, so the amplitude grows too.

Problem 10: Visitors of an Aquarium (xx)

A public aquarium counts its visitors per day. The pattern repeats every week: the most visitors come on Saturdays, 450 per day. The model reaches its lowest value, 110 visitors per day, halfway between Tuesday and Wednesday. The aquarium models the visitors per day by $V(t) = A\cos(B(t - C)) + D$, where $t$ is the day of the week ($t = 0$ Monday, $t = 1$ Tuesday, …, $t = 6$ Sunday). Round numbers of visitors to whole visitors.

Determine $A$, $B$, $C$ and $D$ and give a reason for your choice of $C$.

(for verification: $V(t) = 280 + 170\cos\left(\frac{2\pi}{7}(t - 5)\right)$)

Compute the visitors for every day of the week with your calculator. The indicator R (radians) must show. Explain why Friday and Sunday have the same number of visitors.
On days with more visitors than the midline value, the aquarium employs an extra guide. Determine these days.
Compute the number of visitors per week and the weekly revenue at an entrance fee of 14 EUR. Compare the number of visitors per week with $7 \cdot 280$ and explain the result with the midline.
From next month on, the aquarium moves its feeding show from Saturday to Sunday, and the whole weekly pattern moves one day later. Give the new function $V_1(t)$, name the transformation and determine the days with the fewest visitors.

Solution

a) Amplitude $A = \frac{450 - 110}{2} = 170$, midline $D = \frac{450 + 110}{2} = 280$, period 7 days, so $B = \frac{2\pi}{7}$.

A cosine starts at its maximum. The maximum is on Saturday, $t = 5$, so the cosine is shifted 5 days to the right: $C = 5$.

\[V(t) = 280 + 170\cos\left(\frac{2\pi}{7}(t - 5)\right)\]

Check: the minimum lies half a period (3.5 days) before the maximum, at $t = 1.5$, in the middle between Tuesday and Wednesday ✓

b) Calculator in radian mode:

Day	Mon	Tue	Wed	Thu	Fri	Sat	Sun
$t$	0	1	2	3	4	5	6
$V(t)$	242.17	126.84	126.84	242.17	385.99	450	385.99
Visitors	242	127	127	242	386	450	386

Friday ($t = 4$) and Sunday ($t = 6$) both lie one day away from the maximum on Saturday. The graph is symmetric about its maximum, so both days have the same number of visitors. For the same reason, Tuesday and Wednesday (half a day from the minimum) and Monday and Thursday (1.5 days from the minimum) are equal.

c) Condition: $V(t) > 280 \iff \cos\left(\frac{2\pi}{7}(t - 5)\right) > 0$.

The cosine is positive between its midline crossings at $u = -\frac{\pi}{2}$ and $u = \frac{\pi}{2}$, around its maximum at $u = 0$:

\[-\frac{\pi}{2} < \frac{2\pi}{7}(t - 5) < \frac{\pi}{2} \iff -\frac{7}{4} < t - 5 < \frac{7}{4} \iff 3.25 < t < 6.75\]

The days in this interval are $t = 4, 5, 6$: Friday, Saturday and Sunday. This matches b): 386, 450 and 386 visitors, all other days below 280.

d) Visitors per week: $242 + 127 + 127 + 242 + 386 + 450 + 386 = 1{,}960$. Revenue: $1{,}960 \cdot 14 = 27{,}440$ EUR per week.

$7 \cdot 280 = 1{,}960$, exactly the same. The days above the midline have $106 + 170 + 106 = 382$ visitors more than 280, and the days below have $38 + 153 + 153 + 38 = 382$ fewer. Over a whole period, the deviations above and below the midline cancel out, so on average the aquarium has 280 visitors per day, the midline value.

e) One day later: the visitors of day $t - 1$ now come on day $t$, so $t$ is replaced by $t - 1$. This is a horizontal shift by 1 to the right:

\[V_1(t) = V(t - 1) = 280 + 170\cos\left(\frac{2\pi}{7}(t - 6)\right)\]

The maximum is now on Sunday ($t = 6$), and the minimum half a period earlier, at $t = 2.5$. The fewest visitors now come on Wednesday and Thursday ($t = 2$ and $t = 3$, 127 visitors each).

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: Multi-Storey Car Park in a City Centre [21 pts. total]

A company runs a multi-storey car park (a car park with several floors) with 480 parking spaces in a city centre. The number of occupied spaces follows the same pattern every day. It is highest at 13:00 with 420 cars and lowest at 1:00 with 60 cars. At 7:00, it lies exactly halfway between these two values and rises. The company models the number of occupied spaces by a sinusoidal function $P(t)$, where $t$ is the time in hours after 7:00 ($t = 0$ stands for 7:00, $t = 6$ for 13:00, $0 \leq t < 24$). Round numbers of cars to whole cars and percentages to one decimal place.

Part a)

Determine the amplitude, the midline, the period and the parameter $B$ of the model. [4 pts.]

Solution

Amplitude: half the difference between the highest and the lowest value:

\[A = \frac{420 - 60}{2} = 180 \text{ cars}\]

Midline: the value halfway between them:

\[D = \frac{420 + 60}{2} = 240, \text{ i.e. } y = 240\]

Period: the pattern repeats every day, so the period is 24 hours. Check: from the maximum at 13:00 to the minimum at 1:00 are 12 hours, half a period ✓

Parameter $B$:

\[B = \frac{2\pi}{24} = \frac{\pi}{12}\]

Grading: 1 Pt. amplitude, 1 Pt. midline, 1 Pt. period, 1 Pt. $B$.

Part b)

Decide whether a sine, a cosine or a negative cosine fits the situation at 7:00, and give a reason. Give the function term $P(t)$ and check it with the values at 13:00 and at 1:00. [3 pts.]

(for verification: $P(t) = 240 + 180\sin\left(\frac{\pi}{12}t\right)$)

Solution

Choice: at $t = 0$ (7:00), the number of cars lies on the midline and rises. A sine starts on the midline and rises first. A cosine would start at its maximum, a negative cosine at its minimum.

\[P(t) = 240 + 180\sin\left(\frac{\pi}{12}t\right)\]

Check:

13:00 ($t = 6$): $P(6) = 240 + 180\sin\left(\frac{\pi}{2}\right) = 240 + 180 = 420$ ✓
1:00 ($t = 18$): $P(18) = 240 + 180\sin\left(\frac{3\pi}{2}\right) = 240 - 180 = 60$ ✓

Grading: 1 Pt. sine with reason, 1 Pt. function term, 1 Pt. check at $t = 6$ and $t = 18$.

Part c)

Compute the number of occupied spaces at 9:00 and at 11:00, exactly and rounded. Determine the share of free spaces at 11:00 in percent. [3 pts.]

Solution

9:00 ($t = 2$):

\[P(2) = 240 + 180\sin\left(\frac{\pi}{6}\right) = 240 + 180 \cdot \frac{1}{2} = 330 \text{ cars}\]

11:00 ($t = 4$):

\[P(4) = 240 + 180\sin\left(\frac{\pi}{3}\right) = 240 + 180 \cdot \frac{\sqrt{3}}{2} = 240 + 90\sqrt{3} \approx 395.88, \text{ i.e. about } 396 \text{ cars}\]

Free spaces at 11:00: $480 - 396 = 84$, a share of

\[\frac{84}{480} = 0.175 = 17.5\,\%\]

Grading: 1 Pt. $P(2) = 330$, 1 Pt. $P(4)$ exactly and rounded, 1 Pt. 17.5 % free spaces.

Part d)

Parking costs 1.50 EUR per hour. For every hour, the company has costs of 180 EUR (staff, lighting, lease) plus 0.30 EUR per occupied space (ticket system, cleaning). Give the profit per hour $G(x)$ for $x$ occupied spaces. Determine the break-even point (the number of occupied spaces at which the profit per hour is zero) and the profit per hour at 13:00. [3 pts.]

Solution

Revenue and costs per hour in EUR for $x$ occupied spaces: $R(x) = 1.5x$ and $C(x) = 180 + 0.3x$.

\[G(x) = R(x) - C(x) = 1.5x - (180 + 0.3x) = 1.2x - 180\]

Break-even point:

\[1.2x - 180 = 0 \iff x = \frac{180}{1.2} = 150\]

With more than 150 occupied spaces, the car park makes a profit in that hour. Each car brings $1.50 - 0.30 = 1.20$ EUR towards the costs of 180 EUR.

13:00: $G(420) = 1.2 \cdot 420 - 180 = 504 - 180 = 324$ EUR per hour.

Grading: 1 Pt. $G(x) = 1.2x - 180$, 1 Pt. break-even point 150 cars, 1 Pt. 324 EUR at 13:00.

Part e)

Determine exactly the times of day at which the car park makes a loss. Interpret the result for the company. [4 pts.]

Hint: in $[0; 2\pi]$, $\sin(u) = -\frac{1}{2}$ exactly for $u = \frac{7\pi}{6}$ and $u = \frac{11\pi}{6}$.

Solution

Condition: a loss means fewer than 150 occupied spaces (part d)):

\[240 + 180\sin\left(\frac{\pi}{12}t\right) < 150 \iff 180\sin\left(\frac{\pi}{12}t\right) < -90 \iff \sin\left(\frac{\pi}{12}t\right) < -\frac{1}{2}\]

Boundaries: with $u = \frac{\pi}{12}t$, $t \in [0; 24)$ gives $u \in [0; 2\pi)$. By the hint:

\[\frac{\pi}{12}t = \frac{7\pi}{6} \iff t = 14, \qquad \frac{\pi}{12}t = \frac{11\pi}{6} \iff t = 22\]

Interval: between these values, the sine passes its minimum $-1$ at $u = \frac{3\pi}{2}$ ($t = 18$), so there it is smaller than $-\frac{1}{2}$. Outside, it is larger:

\[\frac{7\pi}{6} < \frac{\pi}{12}t < \frac{11\pi}{6} \iff 14 < t < 22\]

Check: $P(14) = 240 + 180 \cdot \left(-\frac{1}{2}\right) = 150$ ✓ and $P(18) = 60 < 150$ ✓

Interpretation: $t = 14$ is 21:00 and $t = 22$ is 5:00. The car park makes a loss from 21:00 to 5:00, i.e. for 8 hours every night, and a profit from 5:00 to 21:00. The company could reduce its night costs, e.g. with less staff, or close some floors at night.

Grading: 1 Pt. inequality $\sin\left(\frac{\pi}{12}t\right) < -\frac{1}{2}$, 1 Pt. boundaries $t = 14$ and $t = 22$ from the hint, 1 Pt. interval with reason, 1 Pt. clock times and interpretation.

Part f)

The company also runs a car park in a residential area. There, the residents park overnight and drive away in the morning, so the number of occupied spaces is the mirror image of $P(t)$ at its midline. Give a function $P_1(t)$ which is the mirror image of $P(t)$ at the midline and give a reason for your choice. Show that the two car parks together always have 480 occupied spaces, and state at which time the car park in the residential area is fullest. [4 pts.]

Solution

\[P_1(t) = 240 - 180\sin\left(\frac{\pi}{12}t\right)\]

Reason: mirroring at the midline $y = 240$ keeps every $t$ and turns a value $240 + d$ into $240 - d$. As transformations: move the midline to the $t$-axis ($-240$), reflect at the $t$-axis (factor $-1$), move it back ($+240$):

\[P_1(t) = -(P(t) - 240) + 240 = 480 - P(t)\]

The midline $y = 240$ and the amplitude 180 stay, only the sign of the sine term changes. $-P(t)$ would be the mirror image at the $t$-axis, with negative numbers of cars.

Together: $P(t) + P_1(t) = 240 + 240 = 480$ for every $t$ ✓

Fullest: $P_1$ is largest when $\sin\left(\frac{\pi}{12}t\right) = -1$, i.e. $t = 18$. At 1:00, the residential car park has $240 + 180 = 420$ occupied spaces, exactly when the car park in the city centre is emptiest.

Grading: 1 Pt. $P_1(t)$, 1 Pt. reason, 1 Pt. sum 480, 1 Pt. 1:00 with 420 cars.

Calculator Practice

Problem 11: Calculator: Trigonometric Functions (x)

The indicator D (degrees) must show on your calculator. Calculate with the sin, cos and tan keys and round to four decimal places.

$\sin(40°)$
$\cos(75°)$
$\tan(20°)$
$\sin^{-1}(0.8)$ with SHIFT + sin
Check with the sin key that $180° - \theta$ has the same sine as your angle $\theta$ from d). Explain with the unit circle why the calculator shows only one angle in d).

Solution

$\sin(40°) \approx 0.6428$
$\cos(75°) \approx 0.2588$
$\tan(20°) \approx 0.3640$
$\sin^{-1}(0.8) \approx 53.1301°$
$\sin(180° - 53.1301°) = \sin(126.8699°) \approx 0.8000$ ✓. The points for $\theta$ and $180° - \theta$ on the unit circle are mirror images at the $y$-axis, so they have the same $y$-coordinate, the sine. SHIFT + sin gives only the principal value, the angle between $-90°$ and $90°$.

Problem 12: Calculator: Degrees and Radians (x)

Round to four decimal places.

Convert 50° to radians, exactly and rounded.
Convert 2.5 radians to degrees.
With the indicator R (radians): calculate $\cos(2.5)$. With the indicator D: calculate the cosine of your angle from b). Compare.
A student calculates $\sin(2)$, where 2 is an angle in radians, and gets 0.0349. Explain the mistake and give the correct value.

Solution

$50° \cdot \frac{\pi}{180°} = \frac{5\pi}{18} \approx 0.8727$
$2.5 \cdot \frac{180°}{\pi} = \frac{450°}{\pi} \approx 143.2394°$
$\cos(2.5) \approx -0.8011$ (indicator R) and $\cos(143.2394°) \approx -0.8011$ (indicator D): the same angle in two units gives the same value.
The indicator D was showing, so the calculator computed $\sin(2°) \approx 0.0349$. With the indicator R: $\sin(2) \approx 0.9093$.