Define periodic relationship .

A periodic relationship exists between two quantities when, as the input values increase, the output values repeat in a pattern over successive equal-length intervals. The same pattern of outputs occurs again and again, as it does for ocean tides or for the height of a point on a spinning wheel.

How is the graph of a periodic relationship built from a single cycle?

One cycle is copied and the copies are placed end to end , extending in both directions along the input axis. A single cycle is enough because it captures the complete pattern of output values before that pattern starts repeating.

On the graph of a periodic function, between which two points do you measure to find the period?

Any two consecutive points where the same behavior begins , and the period is the horizontal distance between them. Two consecutive peaks will do, or two consecutive troughs, or two consecutive points where the graph crosses the same horizontal level in the same direction.

True or False? If a periodic function is increasing on part of one cycle, it is increasing on the corresponding part of every cycle.

True. Every characteristic found in one period is repeated in every period, so intervals of increase and decrease recur in the same relative positions. The same holds for the concavities of the graph, for the maximum and minimum values, and for the rates of change.

How can the period be found from a table of values at equally spaced inputs?

Read down the output column and find where the values start running through the same sequence again. The period is the difference between the input at the start of one cycle and the input at the start of the next.

Why is a periodic function completely determined by any interval as wide as its period?

Because the outputs repeat every period, so any input outside that interval has the same output as a corresponding input inside it. Knowing what the function does across one full period therefore tells you what it does everywhere.

What does the sign of an angle in standard position tell you?

It gives the direction of the rotation from the initial ray. A positive measure is a rotation counterclockwise , and a negative measure is a rotation clockwise .

Why does the standard position approach work for angles a right triangle cannot represent?

Because it is built from the coordinates of a point on the terminal ray rather than from side lengths, and coordinates may be negative or zero. That covers angles beyond the first quadrant, angles greater than one full revolution and negative angles, none of which fit inside a right triangle.

What is the domain of the sine and cosine functions, and what does that mean for their graphs?

The domain of both is all real numbers , so any real number may be used as an input angle measure. The graphs therefore continue without end in both directions, covering negative angles, which are measured clockwise, and angles carried on past one full revolution.

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Trigonometric Functions (College Board AP® Precalculus): Flashcards

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Define periodic relationship.

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Define periodic relationship.
A periodic relationship exists between two quantities when, as the input values increase, the output values repeat in a pattern over successive equal-length intervals.
The same pattern of outputs occurs again and again, as it does for ocean tides or for the height of a point on a spinning wheel.
How is the graph of a periodic relationship built from a single cycle?
One cycle is copied and the copies are placed end to end, extending in both directions along the input axis.
A single cycle is enough because it captures the complete pattern of output values before that pattern starts repeating.
Fill in the two gaps in the definition of the period of a periodic function.
The period is the $______$ positive value of $k$ for which $f (x + k) =______$ for all $x$ in the domain.
The completed definition is:
The period is the smallest positive value of $k$ for which $f (x + k) = f (x)$ for all $x$ in the domain.
The word smallest is what makes the period unique, since $2 k$ and $3 k$ satisfy the equation just as well.
On the graph of a periodic function, between which two points do you measure to find the period?
Any two consecutive points where the same behavior begins, and the period is the horizontal distance between them.
Two consecutive peaks will do, or two consecutive troughs, or two consecutive points where the graph crosses the same horizontal level in the same direction.
True or False?
If a periodic function is increasing on part of one cycle, it is increasing on the corresponding part of every cycle.
True.
Every characteristic found in one period is repeated in every period, so intervals of increase and decrease recur in the same relative positions.
The same holds for the concavities of the graph, for the maximum and minimum values, and for the rates of change.
How can the period be found from a table of values at equally spaced inputs?
Read down the output column and find where the values start running through the same sequence again.
The period is the difference between the input at the start of one cycle and the input at the start of the next.
Why is a periodic function completely determined by any interval as wide as its period?
Because the outputs repeat every period, so any input outside that interval has the same output as a corresponding input inside it.
Knowing what the function does across one full period therefore tells you what it does everywhere.
Define standard position for an angle.
An angle is in standard position when its vertex is at the origin and one ray, the initial ray, lies along the positive $x$ -axis.
The other ray is the terminal ray, and the measure of the angle is the amount of rotation from the initial ray to the terminal ray.
What does the sign of an angle in standard position tell you?
It gives the direction of the rotation from the initial ray.
A positive measure is a rotation counterclockwise, and a negative measure is a rotation clockwise.
Define coterminal angles.
Two angles in standard position are coterminal when they share the same terminal ray.
So $\frac{π}{4}$ and $\frac{9 π}{4}$ are coterminal, because rotating one further full turn lands on the same ray.
True or False?
An angle in standard position has exactly one coterminal angle.
False.
Coterminal angles differ by a multiple of $2 π$ radians, and any whole number of full revolutions may be added or subtracted, so every angle has infinitely many.
Starting from $\frac{5 π}{3}$ gives $- \frac{π}{3}$ and $- \frac{7 π}{3}$ going one way, and $\frac{11 π}{3}$ going the other.
How is the radian measure of an angle defined?
As the ratio of the arc length the angle subtends to the radius of the circle:
$θ = \frac{arc length}{radius} = \frac{s}{r}$
Both quantities are lengths measured on the same circle, so the ratio is the same whatever size that circle is.
Why does one full revolution measure $2 π$ radians?
On a unit circle the radius is $1$ , so the radian measure of an angle is simply the length of the arc it subtends.
One full revolution subtends the whole circumference of that circle, and the circumference is $2 π$ .
Fill in the two gaps in the rules for converting between degrees and radians.
To convert an angle from degrees to radians, multiply it by $______$ and then simplify. To go from radians back to degrees, multiply by $______$ instead.
The completed rules are:
To convert an angle from degrees to radians, multiply it by $\frac{π}{180}$ and then simplify. To go from radians back to degrees, multiply by $\frac{180}{π}$ instead.
Both factors come from the single relationship $180 ° = π$ radians, written as a fraction one way up or the other.
What pattern connects the multiples of $30 °$ and of $45 °$ to their radian measures?
Multiples of $30 °$ have a denominator of $6$ and multiples of $45 °$ have a denominator of $4$ , with the numerator counting how many multiples there are.
So $150 °$ is the fifth multiple of $30 °$ and becomes $\frac{5 π}{6}$ , though many such fractions then simplify, as $\frac{2 π}{6}$ does to $\frac{π}{3}$ .
How do you find the length of an arc subtended by an angle on a circle of radius $r$ ?
Rearrange $θ = \frac{s}{r}$ to give $s = r θ$ , with the angle measured in radians.
On a circle of radius $4$ , an angle of $\frac{5 π}{3}$ subtends an arc of length $4 \times \frac{5 π}{3} = \frac{20 π}{3}$ .
For an angle $θ$ in standard position, how are sine and cosine defined using a circle of radius $r$ centered at the origin?
The terminal ray meets the circle at a point $P$ with coordinates $(x, y)$ , and the two functions are the ratios of those coordinates to the radius:
$\sin θ = \frac{y}{r}, \cos θ = \frac{x}{r}$
Sine uses the vertical displacement of $P$ from the $x$ -axis, and cosine the horizontal displacement of $P$ from the $y$ -axis.
True or False?
On a circle of radius $7$ , $\sin θ$ is the $y$ -coordinate of the point where the terminal ray meets the circle.
False.
That shortcut holds only on the unit circle, where the radius is $1$ .
On a circle of radius $7$ you must divide by the radius, so $\sin θ = \frac{y}{7}$ and $\cos θ = \frac{x}{7}$ .
How is the tangent of an angle in standard position defined?
It is the slope of the terminal ray, which for a line through the origin is $\tan θ = \frac{y}{x}$ for any point $(x, y)$ on the ray other than the origin.
Equivalently $\tan θ = \frac{\sin θ}{\cos θ}$ , provided $\cos θ \neq 0$ .
An angle $θ$ is in standard position and satisfies $\frac{π}{2} < θ < π$ . Fill in the two gaps.
For this angle $\sin θ$ comes out $______$ while both $\cos θ$ and $\tan θ$ come out $______$ instead.
The completed sentence is:
For this angle $\sin θ$ comes out positive while both $\cos θ$ and $\tan θ$ come out negative instead.
The terminal ray lands where $x < 0$ and $y > 0$ , so $\frac{y}{r}$ is positive while $\frac{x}{r}$ and $\frac{y}{x}$ are both negative.
In the definitions of sine and cosine, why can a displacement be negative when a distance cannot?
Because a displacement records direction as well as size, so a point with a negative $x$ -coordinate has a negative horizontal displacement from the $y$ -axis.
The distance from the origin to the point is a radius and is always positive, which is why the sign of sine or cosine comes entirely from the coordinate.
How do the standard position definitions relate to SOHCAHTOA?
Where $0 < θ < \frac{π}{2}$ , dropping a vertical line from $P$ to the $x$ -axis makes a right triangle with legs $x$ and $y$ and hypotenuse $r$ .
The ratios $\frac{y}{r}$ , $\frac{x}{r}$ and $\frac{y}{x}$ are then exactly opposite over hypotenuse, adjacent over hypotenuse, and opposite over adjacent.
Why does the standard position approach work for angles a right triangle cannot represent?
Because it is built from the coordinates of a point on the terminal ray rather than from side lengths, and coordinates may be negative or zero.
That covers angles beyond the first quadrant, angles greater than one full revolution and negative angles, none of which fit inside a right triangle.
What is the domain of the sine and cosine functions, and what does that mean for their graphs?
The domain of both is all real numbers, so any real number may be used as an input angle measure.
The graphs therefore continue without end in both directions, covering negative angles, which are measured clockwise, and angles carried on past one full revolution.
True or False?
The sine function takes every value between $- 1$ and $1$ .
True.
The range is the whole interval $[- 1, 1]$ , not just the two extreme values.
As the point travels round the unit circle its $y$ -coordinate sweeps through every height in between, and the same is true of cosine.
The exact values of sine and cosine at $\frac{π}{6}$ and at $\frac{π}{4}$ come from two right triangles. Which triangles are they?
An isosceles right triangle with both legs $1$ , whose hypotenuse is $\sqrt{2}$ by Pythagoras, gives $\sin \frac{π}{4} = \cos \frac{π}{4} = \frac{\sqrt{2}}{2}$ .
Half an equilateral triangle of side $2$ , whose third side is $\sqrt{3}$ , gives $\sin \frac{π}{6} = \frac{1}{2}$ and $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ .
An angle of $\frac{π}{2}$ is in standard position. Fill in the two gaps to give the point where its terminal ray meets the unit circle.
$(\cos \frac{π}{2}, \sin \frac{π}{2}) = (______,______)$
The completed coordinates are:
$(\cos \frac{π}{2}, \sin \frac{π}{2}) = (0, 1)$
A quarter turn puts the terminal ray along the positive $y$ -axis, and on the unit circle the coordinates of that point are the cosine and the sine.
The values $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\sin \frac{π}{6} = \frac{1}{2}$ are known. How can the symmetry of the unit circle give $\cos \frac{5 π}{6}$ and $\sin \frac{5 π}{6}$ from them?
Since $\frac{5 π}{6} = π - \frac{π}{6}$ , its point on the unit circle is the reflection over the $y$ -axis of the point for $\frac{π}{6}$ , so the $x$ -coordinate changes sign and the $y$ -coordinate does not.
That gives $\cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2}$ and $\sin \frac{5 π}{6} = \frac{1}{2}$ .
Why do the graphs of sine and cosine repeat after every full revolution?
Because a full revolution brings the terminal ray back to exactly the same position on the unit circle.
The point where it meets the circle then has the same coordinates as before, so both $\cos θ$ and $\sin θ$ return to the values they had $2 π$ earlier.
How does the shape of the sine graph follow from a point moving round the unit circle?
Sine tracks the $y$ -coordinate of that point, so the graph follows it quarter by quarter:
- from $0$ to $\frac{π}{2}$ it rises to the maximum of $1$
- from $\frac{π}{2}$ to $π$ it falls back to $0$
- from $π$ to $\frac{3 π}{2}$ it falls to the minimum of $- 1$
- from $\frac{3 π}{2}$ to $2 π$ it rises back to $0$ , completing one cycle
What is the relationship between the graphs of sine and cosine?
Each is a horizontal translation of the other.
Shifting the sine graph $\frac{π}{2}$ to the left gives the cosine graph, and shifting the cosine graph $\frac{π}{2}$ to the right gives the sine graph.
Why is the period of the tangent function $π$ and not $2 π$ like sine and cosine?
Because $\tan θ$ is the slope of the terminal ray, and the rays for $θ$ and $θ + π$ point in opposite directions but lie on the same line.
The same line has the same slope, so half a revolution is already enough to return the tangent to its previous value.
How do you find the exact value of $\tan \frac{7 π}{6}$ without a calculator?
Use the period first, since tangent repeats every $π$ , so $\tan \frac{7 π}{6} = \tan \frac{π}{6}$ .
Then take the ratio of the known sine and cosine values:
$\tan \frac{π}{6} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
Where does the graph of $y = \tan θ$ have vertical asymptotes, and why?
At $θ = \frac{π}{2} + k π$ for every integer $k$ , because $\cos θ = 0$ at exactly those input values.
There the terminal ray is vertical, so its slope does not exist and the function is undefined.
True or False?
As $θ$ approaches $\frac{π}{2}$ from either side, $\tan θ$ increases without bound.
False.
The two sides behave in opposite ways.
Approaching $\frac{π}{2}$ from the left, $\tan θ$ increases without bound; approaching it from the right, $\tan θ$ decreases without bound.
Describe the behavior of the tangent function between two consecutive asymptotes.
The function is increasing all the way across the interval.
The graph is concave down over the first half and concave up over the second half, changing over at the point of inflection where $\tan θ = 0$ .
True or False?
Between consecutive asymptotes the tangent function takes only output values between $- 1$ and $1$ .
False.
Between any two consecutive asymptotes tangent takes every real number as an output.
The values run from arbitrarily large negative numbers just after one asymptote to arbitrarily large positive numbers just before the next, which is quite unlike the range of sine and cosine.
The function $g$ is given by $g (θ) = a \tan (b (θ + c)) + d$ . Fill in the two gaps in the formula for its period.
$period = \frac{______}{|______|}$
The completed formula is:
$period = \frac{π}{|b|}$
The numerator is $π$ because that is the period of $\tan θ$ itself, and using $2 π$ there is the commonest error, borrowed from the sine and cosine formula.
How can the vertical asymptotes of $g (θ) = a \tan (b (θ + c)) + d$ be found?
Set the whole input of the tangent equal to where $\tan$ is undefined, then solve for $θ$ :
$b (θ + c) = \frac{π}{2} + k π$
For $g (θ) = - 2 \tan \frac{θ}{3} + 4$ this gives $\frac{θ}{3} = \frac{π}{2} + k π$ , so the asymptotes are at $θ = \frac{3 π}{2} + 3 k π$ .
What does the parameter $d$ do to the points of inflection of a tangent graph?
It moves the line containing them from $y = 0$ up to $y = d$ .
The asymptotes stay exactly where they are, so the graph still turns over midway between consecutive asymptotes, just at height $d$ .
The graph of $y = \tan θ$ is increasing between consecutive asymptotes. What does a negative value of $a$ in $y = a \tan θ$ do to that?
It reflects the graph over the $x$ -axis, so between consecutive asymptotes the function is decreasing instead.
The period and the positions of the asymptotes are untouched, because $a$ acts on the output only.

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