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Question 1 of 4
Which of the following graphs represent the trigonometric functions in order from left to right: sin,cos and tan?
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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
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Question 2 of 4
Graph the trigonometric function
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First, identify the values of the sin function
Next, solve for the period of the sin function
Psin |
= |
2πb |
|
= |
2π1 |
Substitute known values |
|
= |
2π |
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 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=4π)
Therefore, this will be the sin curve for y=sin x with a period of 2π
Next, recall that csc x=1sin x
Given this, we can graph y=csc x with regards to the following:
|
= |
Undefined values of 1sin x will be asymptotes |
|
= |
Defined values of 1sin x will be points of intersection |
Asymptotes
1sin 0 |
= |
undefined |
|
1sin π |
= |
undefined |
|
1sin 2π |
= |
undefined |
Points of intersection
1sin π2 |
= |
1 |
|
1sin 3π2 |
= |
-1 |
Take note that the curves of y=csc x will keep approaching but will never intersect with the asymptotes.
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Question 3 of 4
Graph the trigonometric function
y=3sin(x2)
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First, identify the values of the function
Next, solve for the period of the function
Psin |
= |
2πb |
|
= |
2π12 |
Substitute known values |
|
= |
4π |
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 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=4π)
Therefore, this will be the sin curve for y=3sin x2 with a period of 4π
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Question 4 of 4
Graph the trigonometric function within the domain 0 to 2π
y=x+sinx
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Derivatives of Trigonometric Functions
First, get the first and second derivative of the function
y |
= |
x+sin x |
y’ |
= |
1+cos x |
y” |
= |
-sin x |
Next, equate y’=0 to get the stationary point
y’ |
= |
0 |
1+cos x |
= |
0 |
1+cos x -1 |
= |
0 -1 |
Subtract 1 from both sides |
cos x |
= |
-1 |
Recall that cos π=-1. Therefore, x=π
Next, substitute the value of x to the function and solve for y
y |
= |
x+sin x |
|
= |
π+sin π |
Substitute x=π |
|
= |
π+0 |
sin π=0 |
|
= |
π |
Then, substitute the value of x to the second derivative to find the nature of the stationary point
y” |
= |
-sin x |
|
= |
-sin π |
Substitute x=π |
|
= |
-0 |
sin π=0 |
|
= |
0 |
This means that the function has a horizontal point of inflection at the point (π,π)
Given the domain 0 to 2π, π would be halfway from 0 to 2π
Mark the inflection point at (π,π)
Next, substitute x=0 to the function to see how it will look like from the origin
y |
= |
x+sin x |
|
= |
0+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 2π