What is the reliable way to draw a transformed trigonometric graph?

Take the key coordinates of the original curve, apply the change to each one, then draw a smooth curve through the new points. Working point by point is far safer than trying to picture the whole transformation at once.

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Trigonometric Functions (Edexcel A Level Maths: Pure): Flashcards

Exam code: 9MA0

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FrontGraphs of Trigonometric Functions
Define the period of a trigonometric function.

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Cards in this collection (14)

Define the period of a trigonometric function.
The interval after which the graph repeats itself exactly.
$\sin x$ and $\cos x$ have a period of $360 °$ . $\tan x$ has a period of $180 °$ , so it repeats twice as often.
What values can $\sin x$ and $\cos x$ take, and how does $\tan x$ differ?
$\sin x$ and $\cos x$ always lie between $- 1$ and $1$ inclusive.
$\tan x$ is unbounded: it takes every value from $- \infty$ to $+ \infty$ .
What are the key points of $y = \sin x$ between $0 °$ and $360 °$ ?
The five points are $(0 °, 0)$ , $(90 °, 1)$ , $(180 °, 0)$ , $(270 °, - 1)$ and $(360 °, 0)$ .
One point every $90 °$ is enough to sketch the curve. $y = \cos x$ has the same shape but starts at $(0 °, 1)$ .
True or False?
For any angle $x$ , $\cos (- x) = - \cos x$ .
False.
$\cos (- x) = \cos x$ , because the cosine graph is symmetrical about the $y$ -axis.
It is sine that has rotational symmetry about the origin, giving $\sin (- x) = - \sin x$ .
For which values of $x$ is $\tan x$ undefined, and what happens to the graph there?
At every odd multiple of $90 °$ : $\pm 90 °$ , $\pm 270 °$ , $\pm 450 °$ , and so on.
The graph has a vertical asymptote at each one, so the curve breaks into separate branches that run off towards $- \infty$ and $+ \infty$ .
Your calculator gives one solution to $\sin x = - 0.25$ . Why is that not enough?
It returns only the principal value. A trigonometric equation has infinitely many solutions, and the interval you are given usually contains several.
Use the symmetry and periodicity of the graph to find the others.
How does a sketch give you every solution of $\cos x = - \frac{1}{2}$ in a given interval?
Draw the horizontal line $y = - \frac{1}{2}$ across the sketch and read off every crossing point that lies inside the interval.
For $- 180 ° \leq x \leq 360 °$ that gives $x = - 120 °$ , $120 °$ and $240 °$ .
In $y = n \cos x$ , what does $n$ do to the graph, and to the coordinates?
A vertical stretch of scale factor $n$ .
The $x$ coordinates stay the same and the $y$ coordinates are multiplied by $n$ . For $n > 1$ the curve is taller; for $0 < n < 1$ it is squashed towards the $x$ -axis.
In $y = \sin (n x)$ , what does $n$ do to the graph, and to the coordinates?
A horizontal stretch of scale factor $\frac{1}{n}$ .
The $y$ coordinates stay the same and the $x$ coordinates are multiplied by $\frac{1}{n}$ . For $n > 1$ this squashes the curve, so it repeats more often.
True or False?
The graph of $y = \cos (x + 30 °)$ is the graph of $y = \cos x$ moved $30 °$ to the right.
False.
It moves $30 °$ to the left.
A positive $c$ in $\cos (x + c)$ shifts the graph left, which feels backwards to most people. It is $y = \cos (x - 30 °)$ that moves $30 °$ to the right.
A negative $n$ adds a reflection. In which axis, for $y = n \cos x$ and for $y = \sin (n x)$ ?
$y = n \cos x$ reflects in the x-axis, because it is the $y$ values that change sign.
$y = \sin (n x)$ reflects in the y-axis, because it is the $x$ values that change sign.
On $y = \cos x$ there is a point at $(90 °, 0)$ .
On $y = \cos 3 x$ , that point has moved to:
$(______, 0)$
The point has moved to $(30 °, 0)$ .
The $x$ coordinates are multiplied by $\frac{1}{3}$ , and the $y$ coordinates are unchanged.
How many complete cycles does $y = \cos 3 x$ make between $0 °$ and $360 °$ ?
Three.
The period is divided by 3, giving $\frac{360 °}{3} = 120 °$ , and three lots of $120 °$ fit into $360 °$ .
What is the reliable way to draw a transformed trigonometric graph?
Take the key coordinates of the original curve, apply the change to each one, then draw a smooth curve through the new points.
Working point by point is far safer than trying to picture the whole transformation at once.

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