Graph Trigonometric Functions 1

Topics

Try VividMath Premium to unlock full access

Start Free Trial

Lets see your results!

Points

Score

Time

Retry Quiz

Answered
Review

Question 1 of 4

1. Question
Which of the following graphs represent the trigonometric functions in order from left to right: $sin, cos$ and $tan$ ?
- 1.
- 2.
- 3.
- 4.
Hint

Help Video
Correct

Great Work!
Incorrect
Need Text
Play
Current Time 0:00
/
Duration Time 0:00
Remaining Time -0:00
Stream TypeLIVE
Loaded: 0%
Progress: 0%
0:00
Fullscreen
Video Acceleration:
On
Off

00:00
Mute
Playback Rate
1x
2x
1.5x
1.25x
1x
0.75x
0.5x
Subtitles
subtitles off
Captions
captions off
English
Chapters
Chapters

Graphical representation of trigonometric functions can be easily remembered by their corresponding names

Graphing $sin$ functions

The graph for $sin$ function looks like a horizontal letter $S$ whose center is at the point of origin $(0,0)$

The curve is also symmetrical to the x-axis so the $sin$ function is an odd function

Graphing $cos$ functions

The graph for $cos$ function looks like a horizontal letter $C$ whose center intercepts the y-axis at one point

The curve is also symmetrical to the y-axis so the $cos$ function is an even function

Graphing $tan$ function

The graph for $tan$ function looks like a horizontal small letter $t$ whose intersection is at the point of origin $(0,0)$

Notice that the curve from the small letter $t$ goes downward on the left side, and goes upward on the right side

This indicates the direction the curve will be going to

The curve is also symmetrical to the x-axis so the $tan$ function is an odd function

Question 2 of 4

Graph the trigonometric function

$color(red)(y=sinx)$

$y=cscx$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

General Form of a Sin Function

$y=$ $a$

$\text(sin)$ $b$

$x$

Period Fomula

$P_{\text{sin}}=\frac{2\pi}{\color{#00880A}{b}}$

First, identify the values of the

$sin$ function

$y$	$=$	$a$ $\text(sin)$ $b$ $x$

$y$	$=$	$\color{#004ec4}{a}\;\text{sin}\;x$

$a$	$=$	$1$

$b$	$=$	$1$

Next, solve for the period of the

$sin$ function

$P_{\text{sin}}$	$=$	$\frac{2\pi}{\color{#00880A}{b}}$
	$=$	$\frac{2\pi}{\color{#00880A}{1}}$	Substitute known values
	$=$	$2pi$

To graph the

$sin$ curve, it is better to divide the value of its period into for parts and have the curve meet the following conditions

Curve starts at

$(0,0)$

Curve reaches peak of amplitude (

$a$ ) at 1st quarter

Curve intercepts x-axis at 2nd quarter

Curve reaches minimum amplitude at 3rd quarter

Curve starts at x-axis again at the period (

$P=4pi$ )

Therefore, this will be the

$sin$ curve for

$y=\text(sin) x$ with a period of

$2pi$

Next, recall that

$\text(csc) x=1/(\text(sin) x)$

Given this, we can graph

$y=\text(csc) x$ with regards to the following:

	$=$	$\text(Undefined values of) 1/(\text(sin) x) \text(will be asymptotes)$
	$=$	$\text(Defined values of) 1/(\text(sin) x) \text(will be points of intersection)$

Finally, graph

$y=\text(csc) x$

Asymptotes

$1/(\text(sin) 0)$	$=$	$\text(undefined)$

$1/(\text(sin) pi)$	$=$	$\text(undefined)$

$1/(\text(sin) 2pi)$	$=$	$\text(undefined)$

Points of intersection

$1/(\text(sin) pi/2)$	$=$	$1$

$1/(\text(sin) 3pi/2)$	$=$	$-1$

Take note that the curves of

$y=\text(csc) x$ will keep approaching but will never intersect with the asymptotes.

Question 3 of 4

3. Question
Graph the trigonometric function

$y=3sin (x/2)$
- 1.
- 2.
- 3.
- 4.
Hint

Help Video
Correct

Well Done!
Incorrect
Need Text
Play
Current Time 0:00
/
Duration Time 0:00
Remaining Time -0:00
Stream TypeLIVE
Loaded: 0%
Progress: 0%
0:00
Fullscreen
Video Acceleration:
On
Off

00:00
Mute
Playback Rate
1x
2x
1.5x
1.25x
1x
0.75x
0.5x
Subtitles
subtitles off
Captions
captions off
English
Chapters
Chapters

General Form of a Sin Function

$y=$ $a$ $\text(sin)$ $b$ $x$

Period Fomula

$P_{\text{sin}}=\frac{2\pi}{\color{#00880A}{b}}$

First, identify the values of the function

$y$ $=$ $a$ $\text(sin)$ $b$ $x$

$y$ $=$ $\color{#004ec4}{3}\;\text{sin}\;\frac{x}{\color{#00880A}{2}}$

$a$ $=$ $3$

$b$ $=$ $1/2$

Next, solve for the period of the function

$P_{\text{sin}}$ $=$ $\frac{2\pi}{\color{#00880A}{b}}$

$=$ $\frac{2\pi}{\color{#00880A}{\frac{1}{2}}}$ Substitute known values

$=$ $4pi$

To graph the $sin$ curve, it is better to divide the value of its period into for parts and have the curve meet the following conditions

Curve starts at $(0,0)$

Curve reaches peak of amplitude ( $a$ ) at 1st quarter

Curve intercepts x-axis at 2nd quarter

Curve reaches minimum amplitude at 3rd quarter

Curve starts at x-axis again at the period ( $P=4pi$ )

Therefore, this will be the $sin$ curve for $y=3\text(sin) x/2$ with a period of $4pi$
Question 4 of 4

4. Question
Graph the trigonometric function within the domain $0$ to $2pi$

$y=x+sinx$
- 1.
- 2.
- 3.
- 4.
Hint

Help Video
Correct

Nice Job!
Incorrect
Need Text
Play
Current Time 0:00
/
Duration Time 0:00
Remaining Time -0:00
Stream TypeLIVE
Loaded: 0%
Progress: 0%
0:00
Fullscreen
Video Acceleration:
On
Off

00:00
Mute
Playback Rate
1x
2x
1.5x
1.25x
1x
0.75x
0.5x
Subtitles
subtitles off
Captions
captions off
English
Chapters
Chapters

Derivatives of Trigonometric Functions

$y'(\text(sin))=\text(cos)$

$y'(\text(cos))=-\text(sin)$

$y'(\text(tan))=\text(sec)^2$

First, get the first and second derivative of the function

$y$ $=$ $x+\text(sin) x$

$y’$ $=$ $1+\text(cos) x$

$y”$ $=$ $-\text(sin) x$

Next, equate $y’=0$ to get the stationary point

$y’$ $=$ $0$

$1+\text(cos) x$ $=$ $0$

$1+\text(cos) x$ $-1$ $=$ $0$ $-1$ Subtract $1$ from both sides

$\text(cos) x$ $=$ $-1$

Recall that $\text(cos) pi=-1$ . Therefore, $x=pi$

Next, substitute the value of $x$ to the function and solve for $y$

$y$ $=$ $x+\text(sin) x$

$=$ $pi+\text(sin) pi$ Substitute $x=pi$

$=$ $pi+0$ $\text(sin) pi=0$

$=$ $pi$

Then, substitute the value of $x$ to the second derivative to find the nature of the stationary point

$y”$ $=$ $-\text(sin) x$

$=$ $-\text(sin) pi$ Substitute $x=pi$

$=$ $-0$ $\text(sin) pi=0$

$=$ $0$

This means that the function has a horizontal point of inflection at the point $(pi,pi)$

Given the domain $0$ to $2pi$ , $pi$ would be halfway from $0$ to $2pi$

Mark the inflection point at $(pi,pi)$

Next, substitute $x=0$ to the function to see how it will look like from the origin

$y$ $=$ $x+\text(sin) x$

$=$ $0+\text(sin) 0$ Substitute $x=0$

$=$ $0$

The curve from the origin will look like this:

Finally, connect the curve and continue until it reaches the end of the domain, which is $2pi$

Graph Trigonometric Functions 1

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Great Work!

2. Question

Hint

Excellent!

General Form of a Sin Function

Period Fomula

3. Question

Hint

Well Done!

General Form of a Sin Function

Period Fomula

4. Question

Hint

Nice Job!

Derivatives of Trigonometric Functions

Quizzes

$y$	$=$	$a$ $\text(sin)$ $b$ $x$

$y$	$=$	$\color{#004ec4}{3}\;\text{sin}\;\frac{x}{\color{#00880A}{2}}$

$P_{\text{sin}}$	$=$	$\frac{2\pi}{\color{#00880A}{b}}$
	$=$	$\frac{2\pi}{\color{#00880A}{\frac{1}{2}}}$	Substitute known values
	$=$	$4pi$

$y$	$=$	$x+\text(sin) x$
$y’$	$=$	$1+\text(cos) x$
$y”$	$=$	$-\text(sin) x$

$y’$	$=$	$0$
$1+\text(cos) x$	$=$	$0$
$1+\text(cos) x$ $-1$	$=$	$0$ $-1$	Subtract $1$ from both sides
$\text(cos) x$	$=$	$-1$

$y$	$=$	$x+\text(sin) x$
	$=$	$pi+\text(sin) pi$	Substitute $x=pi$
	$=$	$pi+0$	$\text(sin) pi=0$
	$=$	$pi$

$y”$	$=$	$-\text(sin) x$
	$=$	$-\text(sin) pi$	Substitute $x=pi$
	$=$	$-0$	$\text(sin) pi=0$
	$=$	$0$

$y$	$=$	$x+\text(sin) x$
	$=$	$0+\text(sin) 0$	Substitute $x=0$
	$=$	$0$