Trigonometric models questions with answers

Questions

1. Question 1Greatest and least values, and when

   A Ferris wheel turns at a constant speed. The height, $h$ m, of a car above the ground $t$ minutes after the wheel starts turning is given by $h = 28 + 25\sin (20t)^\circ$. Find the least height of the car.

   Show the worked solution

   1. Step 1: sin of any angle is between −1 and 1. Which value makes h least? (Look at the sign in front.)$h$ is least when $\sin (20t)^\circ = -1$ (the number in front of $\sin$ is positive).
   2. Step 2: Put that value into the model$\text{Least } h = 28 + 25(-1)$$= 3$ m
2. Question 2The period, and the value at a time

   The temperature, $T$ °C, in a greenhouse $t$ hours after midnight is given by $T = 28 - 5\cos (15t)^\circ$. Find the temperature when $t = 8.5$, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Put t = 8.5 in$T = 28 - 5\cos (15 \times 8.5)^\circ$$= 28 - 5\cos 127.5^\circ$
   2. Step 2: Calculator in degree mode, then round to 3 significant figures$= 31.0$ °C
3. Question 3When does y take a value?

   A Ferris wheel turns at a constant speed. The height, $h$ m, of a car above the ground $t$ minutes after the wheel starts turning is given by $h = 23 + 20\sin (12t)^\circ$. Find all the values of $t$, for $0 \leq t \leq 30$, for which $h = 33$.

   Show the worked solution

   1. Step 1: Put h = 33 in and make sin the subject$23 + 20\sin (12t)^\circ = 33$$20\sin (12t)^\circ = 33 - 23$$\sin (12t)^\circ = + \dfrac{1}{2}$
   2. Step 2: Let u = 12t: change the range first$\text{Lowest } u = 0^\circ$$\text{Highest } u = 360^\circ$
   3. Step 3: Basic angle: use the positive value$\alpha = \sin^{-1}\left(\dfrac{1}{2}\right)$$= 30^\circ$
   4. Step 4: ASTC (All, Sine, Tan, Cos): in which quadrants does sin have this sign?$\sin u$ is positive in quadrants 1 and 2.
   5. Step 5: List every angle u in the range, quadrant by quadrant$u = 30^\circ$$u = 180^\circ - 30^\circ$$= 150^\circ$
   6. Step 6: Change back to t: undo the substitution$t = \dfrac{u}{12}$$t = 2.5$$t = 12.5$
4. Question 4How long above or below a value

   An object hanging on a spring moves up and down. Its height, $y$ cm, above the floor $t$ seconds after a timer is started is given by $y = 30 + 6\cos (60t)^\circ$. Find the length of time in one complete cycle, $0 \leq t \leq 6$, for which the height of the object is less than $33$ cm.

   Show the worked solution

   1. Step 1: First solve y = 33 in one cycle: make cos the subject$30 + 6\cos (60t)^\circ = 33$$6\cos (60t)^\circ = 33 - 30$$\cos (60t)^\circ = + \dfrac{1}{2}$
   2. Step 2: Let u = 60t, so one cycle is 0° ≤ u ≤ 360°. Basic angle: use the positive value$\alpha = \cos^{-1}\left(\dfrac{1}{2}\right)$$= 60^\circ$
   3. Step 3: ASTC (All, Sine, Tan, Cos): in which quadrants does cos have this sign?$\cos u$ is positive in quadrants 1 and 4.
   4. Step 4: List every angle u in the range, quadrant by quadrant$u = 60^\circ$$u = 360^\circ - 60^\circ$$= 300^\circ$
   5. Step 5: Change back to t: undo the substitution$t = \dfrac{u}{60}$$t = 1$$t = 5$
   6. Step 6: Is y less than 33 between these times or outside them? Try a time in betweenAt $t = 3$, $y = 24$, so $y < 33$ between these times.
   7. Step 7: The length of time: between them, it is the later time take away the earlier; outside them, it is the period (6 seconds) take away that$\text{Between them} = 5 - 1$$= 4$ seconds$\text{Time} = 4$ seconds
5. Question 5Find a, b and c from a graph

   The height, $h$ m, of a car on a Ferris wheel $t$ minutes after the wheel starts turning goes up and down between $5$ m and $95$ m. One complete cycle takes $12$ minutes, and at $t = 0$ the height of the car is at its least. It is modelled by $h = a\cos (bt)^\circ + c$, where $a$, $b$ and $c$ are constants and $b > 0$. Find the value of $a$.

   Show the worked solution

   1. Step 1: The greatest and least values. From the question$\text{Greatest} = 95$$\text{Least} = 5$
   2. Step 2: The amplitude is half the distance from the least to the greatest$\text{Amplitude} = \dfrac{95 - 5}{2}$$= 45$
   3. Step 3: a cos(bt)° starts at its top when a > 0, at its bottom when a < 0At $t = 0$, $h$ is at its bottom, so $a$ is negative.$a = -45$