What does the period of a trigonometric model tell you?

How long the situation takes to return to where it started and begin repeating. For a wheel modelled against time in minutes it is the time for one complete revolution, so knowing a single cycle fixes the whole model.

A LevelMathsCambridge (CIE)Pure 1FlashcardsTrigonometryModelling with Trigonometric Functions

Modelling with Trigonometric Functions (Cambridge (CIE) A Level Maths: Pure 1): Flashcards

Exam code: 9709

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FrontModelling with Trigonometric Functions
A wheel's height is modelled by $H = 50 - 48 \cos (\frac{π}{15} t + 0.3)$ metres. What do the $50$ and the $48$ tell you?

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A wheel's height is modelled by $H = 50 - 48 \cos (\frac{π}{15} t + 0.3)$ metres. What do the $50$ and the $48$ tell you?
The $50$ is the height of the centre of the wheel, and the $48$ is its radius.
The cosine runs between $- 1$ and $1$ , so the height swings $48$ metres either side of $50$ , between $2$ and $98$ metres.
True or False?
The maximum of $y = 15 - 23 \sin 3 x$ occurs when $\sin 3 x = 1$ .
False.
The coefficient is negative, so the largest $y$ comes from the smallest sine, at $\sin 3 x = - 1$ , giving $y = 38$ .
Putting $\sin 3 x = 1$ gives the minimum instead, $y = - 8$ .
For $H = 50 - 48 \cos (\frac{π}{15} t + 0.3)$ , with $t$ in minutes, how do you find when the passenger is first at the top?
Set the cosine to the value that makes $H$ largest, then solve for $t$ .
Here $\frac{π}{15} t + 0.3 = π$ , giving $t = \frac{15}{π} (π - 0.3) = 13.6$ minutes, and the working must be done in radians because the model is written in them.
What does the period of a trigonometric model tell you?
How long the situation takes to return to where it started and begin repeating.
For a wheel modelled against time in minutes it is the time for one complete revolution, so knowing a single cycle fixes the whole model.
Complete the period $T$ of $\cos (q x + r)$ in each measure:
$T = \frac{______}{q} in degrees, and T = \frac{______}{q} in radians$
The completed formulae are:
$T = \frac{360 °}{q} in degrees, and T = \frac{2 π}{q} in radians$
In both, $T$ comes out in the same units as $x$ , so a model in minutes gives a period in minutes.
What is the period of $y = \cos (\frac{7}{8} x - 1.5)$ , with $x$ in radians?
The period is $\frac{2 π}{\frac{7}{8}} = \frac{16 π}{7}$ .
The $- 1.5$ shifts the curve sideways but leaves the length of one cycle completely untouched.

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