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Question 1 of 4
1. Question
Graph the trigonometric function`y=4cos 3(x+pi/3)`Hint
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General Form of a Cos Function with Horizontal Shift
`y=``a` `\text(cos)` `b``(x``h``)`Period Fomula
$$P=\frac{2\pi}{\color{#00880A}{b}}$$First, identify the values of the function`y` `=` `a` `\text(cos)` `b``(x``h``)` `y` `=` $$\color{#004ec4}{4}\;\text{cos}\;\color{#00880A}{3}(x+\color{#9a00c7}{\frac{\pi}{3}})$$ `a` `=` `4` `b` `=` `3` `h` `=` `pi/3` Next, solve for the period of the function$$P$$ `=` $$\frac{2\pi}{\color{#00880A}{b}}$$ `=` $$\frac{2\pi}{\color{#00880A}{3}}$$ Substitute known values To graph the `cos` curve, it is better to divide the value of its period into `4` parts and have the curve meet the following conditionsCurve starts at the peak of the amplitude `(a=4)`Curve intercepts xaxis at 1st quarterCurve reaches minimum amplitude at 2nd quarterCurve intercepts xaxis again at 3rd quarterCurve starts again at the period (`P=(2pi)/3`)This will be the `cos` curve for `y=4\text(cos) 3x` with a period of `(2pi)/3`Shift the graph horizontally `h=pi/3` units to the left. Since the labels are in intervals of `pi/6`, we move the graph `2` labels to the left 
Question 2 of 4
2. Question
Graph the trigonometric function within the domain `0≤x≤2pi``y=cos x+1`Hint
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General Form of a Cos Function with Vertical Shift
`y=``a` `\text(cos)` `b``x+``k`Period Fomula
$$P=\frac{2\pi}{\color{#00880A}{b}}$$First, identify the values of the function`y` `=` `a` `\text(cos)` `b``x+``k` `y` `=` $$\color{#004ec4}{}\;\text{cos}\;x+\color{#e65021}{1}$$ `a` `=` `1` `b` `=` `1` `k` `=` `1` Next, solve for the period of the function$$P$$ `=` $$\frac{2\pi}{\color{#00880A}{b}}$$ `=` $$\frac{2\pi}{\color{#00880A}{1}}$$ Substitute known values `=` `2pi` To graph the `cos` curve, it is better to divide the value of its period into `4` parts and have the curve meet the following conditionsCurve starts at the minimum amplitude `(a=1)`Curve intercepts xaxis at 1st quarterCurve reaches peak of amplitude at 2nd quarterCurve intercepts xaxis again at 3rd quarterCurve starts again at the period (`P=2pi`)This will be the `cos` curve for `y=\text(cos) x` with a period of `2pi`Finally, move the curve up to match the recently plotted points 
Question 3 of 4
3. Question
Graph the following trigonometric functions within the domain `0≤x≤pi``color(red)(y=cos2x)``color(blue)(y=x/2)`Hint
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General Form of a Cos Function
`y=``a` `\text(cos)` `b``x`Period Fomula
$$P=\frac{2\pi}{\color{#00880A}{b}}$$Start by graphing the first functionFirst, identify the values of the function.`y` `=` `a` `\text(sin)` `b``x` `y` `=` $$\text{cos}\;2x$$ `a` `=` `1` `b` `=` `2` Next, solve for the period of the function$$P$$ `=` $$\frac{2\pi}{\color{#00880A}{b}}$$ `=` $$\frac{2\pi}{\color{#00880A}{2}}$$ Substitute known values `=` `pi` Graph the `1`st function, `y=cos2x`. It is better to divide the value of its period into `4` parts and have the curve meet the following conditions:Curve starts at `(0,1)`Curve intercepts xaxis at 1st quarterCurve reaches minimum amplitude (`a=1`) at 2nd quarterCurve intercepts xaxis at 3rd quarterCurve starts again at the period (`pi,1`)Hence, this will be the curve for `y=cos2x` under the domain `0≤x≤pi`Now, graph the `2`nd functionConvert the current labels in the graph into decimal form and insert whole numbers accordinglySet up a grid of `x` and `y` values to help with the graphingSubstitute each `x` value to function to solve for the corresponding `y` value`x` `0` `1` `2` `y` `y` `=` `x/2` `=` `0/2` Substitute `x=0` `=` `0` `x` `0` `1` `2` `y` `0` `y` `=` `x/2` `=` `1/2` Substitute `x=1` `x` `0` `1` `2` `y` `0` `1/2` `y` `=` `x/2` `=` `2/2` Substitute `x=2` `=` `1` `x` `0` `1` `2` `y` `0` `1/2` `1` Plot the `3` points on the updated graphFinally, connect the `3` dots to form a line 
Question 4 of 4
4. Question
Graph the following trigonometric functions within the domain `2pi≤x≤2pi``y=2sinx``y=2sinx`Hint
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General Form of a Sin Function
`y=``a` `\text(sin)` `b``x`Period Fomula
$$P=\frac{2\pi}{\color{#00880A}{b}}$$First, identify the values of the function`1`st function:`y` `=` `a` `\text(sin)` `b``x` `y` `=` $$\color{#004ec4}{2}\;\text{sin}\;x$$ `a` `=` `2` `b` `=` `1` `2`nd function:`y` `=` `a` `\text(sin)` `b``x` `y` `=` $$\color{#004ec4}{2}\;\text{sin}\;x$$ `a` `=` `2` `b` `=` `1` Next, solve for the period of the function`1`st Function:$$P_{\text{sin}}$$ `=` $$\frac{2\pi}{\color{#00880A}{b}}$$ `=` $$\frac{2\pi}{\color{#00880A}{1}}$$ Substitute known values `=` `2pi` `2`st Function:$$P_{\text{sin}}$$ `=` $$\frac{2\pi}{\color{#00880A}{b}}$$ `=` $$\frac{2\pi}{\color{#00880A}{1}}$$ Substitute known values `=` `2pi` The period of both functions is `2pi`Graph the `1`st function, `y=2sinx`. It is better to divide the value of its period into `4` parts and have the curve meet the following conditions:Curve starts at `(0,0)`Curve reaches peak of amplitude (`a=2`) at 1st quarterCurve intercepts xaxis at 2nd quarterCurve reaches minimum amplitude at 3rd quarterCurve starts at xaxis again at the period (`P=2pi`)Since the domain is `2pi≤x≤2pi`, copy this curve into the left side of the `y`axisHence, this will be the curve for `y=2\text(sin) x` under the domain `2pi≤x≤2pi`Lastly, graph the `2`nd function, `y=2sinx`. Again, divide the value of its period into `4` parts and have the curve meet the following conditions:Curve starts at `(0,0)`Curve reaches peak of amplitude (`a=2`) at 1st quarterCurve intercepts xaxis at 2nd quarterCurve reaches minimum amplitude at 3rd quarterCurve starts at xaxis again at the period (`P=2pi`)Since the domain is `2pi≤x≤2pi`, copy this curve into the left side of the `y`axisTherefore, this will be the curve for the functions `y=2\text(sin) x` and `y=2\text(sin) x` under the domain `2pi≤x≤2pi`
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