a) $y=3sin(2x)+1$

From the graph,.we can observe that

$y_{min}=-2$ and $y_{max}=4$

equation of the axis: $y=dfrac{4+(-2)}{2}implies y=1$

amplitude: $A=dfrac{4-(-2)}{2}=3$

period: $225^circ-45^circ=180^circ$

b) $y=4cos(0.5x)-2$

From the graph,.we can observe that

$y_{min}=-6$ and $y_{max}=2$

equation of the axis: $y=dfrac{2+(-6)}{2}implies y=-2$

amplitude: $A=dfrac{2-(-6)}{2}=4$

period: $720^circ-0^circ=720^circ$

This exercise requires you to use your calculator to evaluate the given functions. Be sure to set it in DEGREE mode.

a.) $h(x)=sin(5x)-1$ when $x=25^circ$

$$
h(x)=sin(5cdot 25)-1approx -0.1808
$$

b) $f(x)=cos x$ such that $f(x)=0$

We can plot $y=cos x$ and see which values of $x$ does the graph touches the line $y=0$ (x-axis).

Thus, within the given interval, $x=90^circ,;270^circ$

a) $-0$.1808

b) $90^circ$, $270^circ$

a) Use your graphing calculator or online graphing tool to graph the given function

$$
h(t)=cos(36^circ t)
$$

b) From the graph, the distance between two peaks is the period which is about $10$ s.

c) At $t=35$, the cycle has gone 3 times plus additional 5 s. Thus, it is the same as when $t=5$ where the displacement is $h=-1$

d) We can approximate from the graph at $t=4$s, the displacement is $-0.8$m.

a) see graph

b) 10 s

c) $h=-1$m

d) $t=4$s

You can visualize the problem by sketching a diagram shown below.

We need to find the coordinates of the point after rotating from the point $(2,0)$ counterclockwise with respect to the origin.

This is basically obtaining the point $(2cos 50^circ,2sin 50^circ)$.

Using your calculator, you should get $(1.286,1.532)$.

$$
(1.286,1.532)
$$

a) $y=3sin x +1$

Periodic graphs are those that repeats at regular intervals.

Sinusoidal graphs are periodic graphs that are smooth and symmetrical waves created by transforming $y=cos x$ or $y=sin x$

a) The graph is periodic and sinusoidal.

b) $y=(0.004x)sin x$

c) $y=cos 2x -sin x$

d) $y=0.005x+sin x$

e) $y=0.5cos x -1$

f) $y=sin 90^circ$

f) This does not look like periodic function but if you remember the definition, it actually is.

By strict definition, a function $f(x)$ is periodic if there exists a non-zero period of $T$ such that

$f(x)=f(x+nT)$ where $n=$ non-zero integer

This property is satisfied by a horizontal line where $T$ could take any real non-zero value.

However, it is not sinusoidal because it is not composed of smooth repeating symmetrical curves.

Given that $g(x)=sin x$ and $h(x)=cos x$ where $0^circ leq x leq 360^circ$

a) $g(90^circ)=sin 90^circ=1$

This means that if $(x,y)$ is a point on the unit circle, rotating counterclockwise by $90^circ$ from the positive $x$-axis would end up to a point with $y$-coordinate of 1.

b) $h(90^circ)=cos 90^circ=0$

This means that if $(x,y)$ is a point on the unit circle, rotating counterclockwise by $90^circ$ from the positive $x$-axis would end up to a point with $x$-coordinate of 0.

Thus, the resulting point is $(0,1)$ when rotated counterclockwise by $90^circ$ from positive $x$-axis.

The general forms of sinusoidal functions can be written as\\
$y=Asin (kx)+B$ or $y=Acos (kx)+B$\\
where\\
amplitude = $A$\\
period = $T$ = $dfrac{360^circ}{k}$ \\
equation of the axis: $y=B$\\
Increasing and decreasing intervals:\\
[begin{gathered}
{text{sine function}} hfill \
{text{inceasing: }} – frac{{{{90}^ circ }}}{k} + Tn leq x leq frac{{{{90}^ circ }}}{k} + Tn hfill \
{text{decreasing: }}frac{{{text{9}}{{text{0}}^ circ }}}{k} + Tn leq x leq frac{{{{270}^ circ }}}{k} + Tn hfill \
hfill \
{text{cosine function}} hfill \
{text{increasing: }}frac{{{text{18}}{{text{0}}^ circ }}}{k} + Tn leq x leq frac{{{{360}^ circ }}}{k} + Tn hfill \
{text{decreasing: }}{{text{0}}^ circ } + Tn,,, leq x, leq ,,,frac{{{{180}^ circ }}}{k} + Tn hfill \
end{gathered} ]\\
where $T$ is the period, $n$ is an integer.\\

a) $y=2sin x +3$

a)

amplitude = $A$ = 2

period = $360^circ$

equation of the axis: $y=3$

increasing at: $-90^circ+360^circ n leq x leq 90 + 360^circ n$

decreasing at: $90^circ +360^circ n leq x leq 270^circ+360^circ n$

where $n$ is an integer.

b) $y=3sin x +1$

b)

amplitude = $A=3$

period = $360^circ$

equation of the axis: $y=1$

increasing at: $-90^circ+360^circ n leq x leq 90 + 360^circ n$

decreasing at: $90^circ +360^circ n leq x leq 270^circ+360^circ n$

where $n$ is an integer.

c) $y=sin (0.5 x) +2$

c)

amplitude = $A=1$

period = $T=dfrac{360}{0.5}=720$

equation of the axis: $y=2$

increasing at: $-180^circ+720^circ n leq x leq 180 + 720^circ n$

decreasing at: $180^circ +720^circ n leq x leq 540^circ+720^circ n$

where $n$ is an integer.

d) $y=sin (2x)-1$

d)

amplitude = $A=1$

period = $T=dfrac{360}{2}=180$

equation of the axis: $y=-1$

increasing at: $-45^circ+180^circ n leq x leq 180 + 720^circ n$

decreasing at: $45^circ +180^circ n leq x leq 135^circ+180^circ n$

where $n$ is an integer.

f) $y=3sin (0.5 x)+2$

f)

amplitude = $A=3$

period = $T=dfrac{360}{0.5}=720$

equation of the axis: $y=3$

increasing at: $-180^circ+720^circ n leq x leq 180 + 720^circ n$

decreasing at: $180^circ +720^circ n leq x leq 540^circ+720^circ n$

where $n$ is an integer.

This exercise requires you to use your calculator to evaluate the given functions. Be sure to set it in DEGREE mode.

a.) $f(x)=cos ximplies f(35^circ)=cos 35^circ approx 0.8192$

b) $g(x)=sin(2x)implies g(10^circ)=sin(2cdot 10^circ)approx 0.3420$

c) $h(x)=cos(3x)+1=cos(3cdot 20^circ)+1=1.5$

d) $f(x)=cos x$ and $f(x)=-1$ within $0^circ leq x leq 360^circ$

From the graph, we see that $x=180^circ$

d) $f(x)=sin x$ and $f(x)=-1$ within $0^circ leq x leq 360^circ$

From the graph, we see that $x=270^circ$

a) $0.8192$

b) $0.3420$

c) 1.5

d) $x=180^circ$

e) $x=270^circ$

You can graph both $y=cos x$ and $y=sin x$ within the given interval in your graphing calculator or online graphing tool. Be sure to set it in DEGREE mode. Find the intersections, the coordinate of $x$ is the solution.

Therefore, $x=-315^circ,;-135^circ, ;45^circ, ;225^circ$

$$
x=-315^circ,;-135^circ, ;45^circ, ;225^circ
$$

The coordinates of the new point $(x,y)$ after rotation of $theta ^circ$ about (0,0) from the point $(r,0)$ is

$(x,y)implies (rcostheta, rsin theta)$

In this exercise, we are given with $r$ and $theta$, so we will simply substitute it as follows.

a) $r=1$, $theta=25^circ$ $implies (1cdot cos 25^circ, 1cdot sin 25^circ)implies (0.9063,0.4226)$

b) $r=5$ , $theta=80^circ$ $implies (5cdot cos 80^circ, 5cdot sin 80^circ) implies (0.8682,4.9240)$

c) $r=4$ , $theta=120^circ$ $implies (4cdot cos 120^circ, 4cdot sin 120^circ) implies (-2,2sqrt{3})$

d) $r=3$ , $theta=230^circ$ $implies (3cdot cos 230^circ, 3cdot sin 230^circ) implies (-1.9284, -2.2981)$

a) $(0.9063,0.4226)$

b) $(0.8682,4.9240)$

c) $(-2,2sqrt{3})$

d) $(-1.9284,-2.2981)$

The general forms of sinusoidal functions can be written as

$y=Asin (kx)+B$

or

$y=Acos (kx)+B$

where

amplitude = $A$

period = $T$ = $dfrac{360^circ}{k}$

equation of the axis: $y=B$

To graph $j$ cycles starting from $x=x_0$, we must end at $x=x_0+Tcdot j$

Here, we can choose either sine or cosine. We shall sine in these examples starting from $x=0$. The sinusoidal function is plotted in red while the equation of the axis is in dashed purple.

a) $T=4$ , $A=3$ , $y=5 implies y=3sin left(dfrac{360^circ}{4}xright)+5$

2 cycles $implies 0leq xleq 8$

b) $T=20$ , $A=6$ , $y=4 implies y=6sin left(dfrac{360^circ}{20}xright)+4$

3 cycles $implies 0leq xleq 60$

b) $T=80$ , $A=5$ , $y=-2 implies y=5sin left(dfrac{360^circ}{80}xright)-2$

2 cycles $implies 0leq xleq 160$

The Ferris wheel can be treated as a circle with radius $r=5$ and center $(0,0)$.

a) $h(t)=5cos (18t)^circ$ represents the horizontal distance between Jim and the center at any time $t$.

Thus, $h(10)=5cos(-180)=-5$ means that Jim is 5 units to the left of the center at 10 s.

b) Similarly, $h(t)=5sin(18t)^circ$ represents the vertical distance from the center. Thus, $h(10)=5sin (180^circ)$ is at the same height as the center of the Ferris wheel at 10s.

The graph of $y=sin x$ and $y=cos x$ is shown below within $0^circ leq x leq 360^circ$

$bf{Similarities}$

Both are periodic and sinusoidal function with the same amplitude $(A=1)$, period $(T=360^circ)$, and axis $(y=0)$.
Both contains the point $(45^circ,0.707)$ and $(225^circ, -0.707)$

$bf{Differences}$

The graph of $y=sin x$ passes through the $y$-axis at the origin $(0,0)$ while $y=cos x$ passes through $(0,1)$. The graph of $y=cos x$ is horizontally translated by $90^circ$ from the graph of $y=sin x$.

The diameter of the wheel is $2$m so when $3/4$ of it exposed, 1/4 of 2 m, which is 0.5 m must be submerged in water.
You can visualize the problem by sketching the figure as follows.

Now, we want to know the function describing the height of the nail in terms of the rotation $theta$.

Remember that $y=sin theta$ represents the vertical distance of the nail from the center of the wheel. Since the center is now shifted $0.5$ m above the water level, the function describing the nails height from the water level is

$$
y=sin theta+0.5
$$

$$
y=sin theta +0.5
$$