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Graphing Trigonometric Functions 1Graphing Trigonometric Functions 1
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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`?Hint
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Graphical representation of trigonometric functions can be easily remembered by their corresponding namesGraphing `sin` functionsThe 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 functionGraphing `cos` functionsThe 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 functionGraphing `tan` functionThe 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 sideThis 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
2. Question
Graph the trigonometric function`color(red)(y=sinx)``y=cscx`Hint
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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 conditionsCurve 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.
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Question 3 of 4
3. Question
Graph the trigonometric function`y=3sin (x/2)`Hint
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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 conditionsCurve 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`
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Question 4 of 4
4. Question
Graph the trigonometric function within the domain `0` to `2pi``y=x+sinx`Hint
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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`
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