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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 xaxis so the `sin` function is an odd functionGraphing `cos` functionsThe graph for `cos` function looks like a horizontal letter `C` whose center intercepts the yaxis at one pointThe curve is also symmetrical to the yaxis 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 toThe curve is also symmetrical to the xaxis so the `tan` function is an odd function 
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 quarterCurve intercepts xaxis at 2nd quarterCurve reaches minimum amplitude at 3rd quarterCurve starts at xaxis 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)`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 quarterCurve intercepts xaxis at 2nd quarterCurve reaches minimum amplitude at 3rd quarterCurve starts at xaxis 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`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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