IB Maths AA HL 3.7 Trigonometric functions Question Bank

Practise IB Mathematics SL/HL 3.7 by applying trigonometric functions methods to exam-style questions.

[Maximum number: 10]

Sule Skerry and Rockall are small islands in the Atlantic Ocean, in the same time zone.
On a given day, the height of water in metres at Sule Skerry is modelled by the function $H(t)=1.63 \sin (0.513(t-8.20))+2.13$ , where t is the number of hours after midnight.
The following graph shows the height of the water for 15 hours, starting at midnight.
At low tide the height of the water is 0.50 m . At high tide the height of the water is 3.76 m .
All heights are given correct to two decimal places.

Figure for Question SL 3.7—Trigonometric functions question 1 — IB Maths AA HL

(a)

The length of time between the first low tide and the first high tide is 6 hours and m minutes. Find the value of m to the nearest integer.

[ 3 ]

EITHER
attempt to find value of t for the first low tide OR the first high tide
$11.2619 \ldots-5.13801 \ldots$ $=6.12396 \ldots$

OR
attempt to find half of the period
$\frac{1}{2} \times \frac{2 \pi}{0.513}$ $=6.12396 \ldots$

THEN
$m=(6.12396 \ldots-6) \times 60=7.43773 \ldots$
m=7

(b)

Find the values of a, b, c and d.

When t=T, the height of the water at Sule Skerry is the same as the height of the water at Rockall for the first time.

[ 7 ]

Note: In part (d), award the marks for a, b, c and d independent of each other.

METHOD 1

\begin{array}{ll} a=1.17 & \text { A1 } \\ d=1.57 & \text { A1 } \end{array}

attempt to find time between low and high tide in hours

6 hours and 21 minutes =6.35 hours
(period =) 12.7

\begin{aligned} & b=\frac{2 \pi}{12.7}=0.494739 \ldots \\ & b=0.495\left(=\frac{60 \pi}{381}\right) \end{aligned}

attempt to find mean of low and high tide times OR substitute values of a known point

\begin{aligned} & c=\frac{1}{2}\left(2 \frac{41}{60}+9 \frac{2}{60}\right) \text { OR eg } 0.40=1.17 \sin (0.495(2.68333 \ldots-c))+1.57 \\ & c=5.85833 \ldots \\ & c=5.86 \end{aligned}

Note: Award (M1)A1 for c=18.6.
Award (M1)A0 for c=-6.84.
METHOD 2
a=1.17
d=1.57
substituting at least one point into h(t)
$1.17 \sin \left(b\left(2 \frac{41}{60}-c\right)\right)+1.57=0.4$ OR $1.17 \sin \left(b\left(9 \frac{2}{60}-c\right)\right)+1.57=2.74$ $b\left(2 \frac{41}{60}-c\right)=-\frac{\pi}{2}(=-1.57)$ AND $b\left(9 \frac{2}{60}-c\right)=\frac{\pi}{2}(=1.57)$

Note: accept any angles of the form $-\frac{\pi}{2}+c \pi k$ and $\frac{\pi}{2}+c \pi k$ .

EITHER
use of graph or table to find their intersection

OR
attempt to solve their equations simultaneously
$\frac{2 \frac{41}{60}-c}{9 \frac{2}{60}-c}=-1$

THEN
$c=5.85833 \ldots$
c=5.86
$b=0.494739 \ldots$
b=0.495

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SL 3.7—Trigonometric functions question 1

Question (a)

Question (b)