Chapter 6

The sine and cosine functions have the property that $f\left(x+P\right)=f\left(x\right)$ for a certain $P.$ This means that the function values repeat for every $P$ units on the x-axis.

The absolute value of the constant $A$ (amplitude) increases the total range and the constant $D$ (vertical shift) shifts the graph vertically.

At the point where the terminal side of $t$ intersects the unit circle, you can determine that the $\sin t$ equals the y-coordinate of the point.

amplitude: 2; midline: $y=-3;$ period: 4; equation: $f(x)=2\sin \left(\frac{\pi }{2}x\right)-3$

amplitude: 2; period: 5; midline: $y=3;$ equation: $f(x)=-2\cos \left(\frac{2\pi }{5}x\right)+3$

amplitude: 4; period: 2; midline: $y=0;$ equation: $f(x)=-4\cos \left(\pi \left(x-\frac{\pi }{2}\right)\right)$

amplitude: 2; period: 2; midline $y=1;$ equation: $f\left(x\right)=2\cos \left(\pi x\right)+1$

$0,\mathrm{\pi }$

$\text{sin}\left(\frac{\mathrm{\pi }}{2}\right)=1$

$\frac{\mathrm{\pi }}{2}$

$f\left(x\right)=\text{sin}x$ is symmetric

$\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}$

Maximum: $1$ at $x=0$ ; minimum: $-1$ at $x=\mathrm{\pi }$

A linear function is added to a periodic sine function. The graph does not have an amplitude because as the linear function increases without bound the combined function $h\left(x\right)=x+\text{sin}x$ will increase without bound as well. The graph is bounded between the graphs of $y=x+1$ and $y=x-1$ because sine oscillates between −1 and 1.

There is no amplitude because the function is not bounded.

The graph is symmetric with respect to the y-axis and there is no amplitude because the function’s bounds decrease as $\left|x\right|$ grows. There appears to be a horizontal asymptote at $y=0$ .

Since $y=\csc x$ is the reciprocal function of $y=\sin x,$ you can plot the reciprocal of the coordinates on the graph of $y=\sin x$ to obtain the y-coordinates of $y=\csc x.$ The x-intercepts of the graph $y=\sin x$ are the vertical asymptotes for the graph of $y=\csc x.$

Answers will vary. Using the unit circle, one can show that $\tan \left(x+\pi \right)=\tan x.$

The period is the same: $2\pi .$

IV

III

period: 8; horizontal shift: 1 unit to left

1.5

5

$-\cot x\cos x-\sin x$

$y=\tan \left(3\left(x-\frac{\pi }{4}\right)\right)+2$

$f\left(x\right)=\csc \left(2x\right)$

$f\left(x\right)=\csc \left(4x\right)$

$f\left(x\right)=2\csc x$

$f(x)=\frac{1}{2}\tan (100\pi x)$

The function $y=\sin x$ is one-to-one on $\left[-\frac{\pi }{2},\frac{\pi }{2}\right];$ thus, this interval is the range of the inverse function of $y=\sin x,$ $f(x)={\sin }^{-1}x.$ The function $y=\cos x$ is one-to-one on $\left[0,\pi \right];$ thus, this interval is the range of the inverse function of $y=\cos x,f(x)={\cos }^{-1}x.$

$\frac{\pi }{6}$ is the radian measure of an angle between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$ whose sine is 0.5.

In order for any function to have an inverse, the function must be one-to-one and must pass the horizontal line test. The regular sine function is not one-to-one unless its domain is restricted in some way. Mathematicians have agreed to restrict the sine function to the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ so that it is one-to-one and possesses an inverse.

True . The angle, ${\theta }_1$ that equals $\arccos (-x)$, $x>0$, will be a second quadrant angle with reference angle, ${\theta }_2$, where ${\theta }_2$ equals $\arccos x$, $x>0$. Since ${\theta }_2$ is the reference angle for ${\theta }_1$, ${\theta }_2=\pi -{\theta }_1$ and $\arccos (-x)$ = $\pi -\arccos x$-

$-\frac{\pi }{6}$

$\frac{3\pi }{4}$

$-\frac{\pi }{3}$

$\frac{\pi }{3}$

1.98

0.93

1.41

0.56 radians

0

0.71

-0.71

$-\frac{\pi }{4}$

0.8

$\frac{5}{13}$

$\frac{x-1}{\sqrt{-x^2+2x}}$

$\frac{\sqrt{x^2-1}}{x}$

$\frac{x+0.5}{\sqrt{-x^2-x+\frac{3}{4}}}$

$\frac{\sqrt{2x+1}}{x+1}$

$\frac{\sqrt{2x+1}}{x}$

$t$

approximately $x=0.00$

0.395 radians

1.11 radians

1.25 radians

0.405 radians

No. The angle the ladder makes with the horizontal is 60 degrees.

amplitude: 3; period: $2\pi ;$ midline: $y=3;$ no asymptotes

amplitude: 3; period: $2\pi ;$ midline: $y=0;$ no asymptotes

amplitude: 3; period: $2\pi ;$ midline: $y=-4;$ no asymptotes

amplitude: 6; period: $\frac{2\pi }{3};$ midline: $y=-1;$ no asymptotes

stretching factor: none; period: $\pi ;$ midline: $y=-4;$ asymptotes: $x=\frac{\pi }{2}+\pi k,$ where $k$ is an integer

stretching factor: 3; period: $\frac{\pi }{4};$ midline: $y=-2;$ asymptotes: $x=\frac{\pi }{8}+\frac{\pi }{4}k,$ where $k$ is an integer

amplitude: none; period: $2\pi ;$ no phase shift; asymptotes: $x=\frac{\pi }{2}k,$ where $k$ is an odd integer

amplitude: none; period: $\frac{2\pi }{5};$ no phase shift; asymptotes: $x=\frac{\pi }{5}k,$ where $k$ is an integer

amplitude: none; period: $4\pi ;$ no phase shift; asymptotes: $x=2\pi k,$ where $k$ is an integer

largest: 20,000; smallest: 4,000

amplitude: 8,000; period: 10; phase shift: 0

In 2007, the predicted population is 4,413. In 2010, the population will be 11,924.

5 in.

10 seconds

$\frac{\pi }{6}$

$\frac{\pi }{4}$

$\frac{\pi }{3}$

No solution

$\frac{12}{5}$

The graphs are not symmetrical with respect to the line $y=x.$ They are symmetrical with respect to the $y$-axis.

The graphs appear to be identical.

amplitude: 0.5; period: $2\pi ;$ midline $y=0$

amplitude: 5; period: $2\pi ;$ midline: $y=0$

amplitude: 1; period: $2\pi ;$ midline: $y=1$

amplitude: 3; period: $6\pi ;$ midline: $y=0$

amplitude: none; period: $\pi ;$ midline: $y=0,$ asymptotes: $x=\frac{2\pi }{3}+\pi k,$ where $k$ is an integer

amplitude: none; period: $\frac{2\pi }{3};$ midline: $y=0,$ asymptotes: $x=\frac{\pi }{3}k,$ where $k$ is an integer

amplitude: none; period: $2\pi ;$ midline: $y=-3$

amplitude: 2; period: 2; midline: $y=0;$ $f\left(x\right)=2\sin \left(\pi \left(x-1\right)\right)$

amplitude: 1; period: 12; phase shift: $\mathrm{-6};$ midline $y=\mathrm{-3}$

$D\left(t\right)=68-12\sin \left(\frac{\pi }{12}x\right)$

period: $\frac{\pi }{6};$ horizontal shift: $\mathrm{-7}$

$f\left(x\right)=\sec \left(\pi x\right);$ period: 2; phase shift: 0

$4$

The views are different because the period of the wave is $\frac{1}{25}.$ Over a bigger domain, there will be more cycles of the graph.

$\frac{3}{5}$

On the approximate intervals $\left(0.5,1\right),\left(1.6,2.1\right),\left(2.6,3.1\right),\left(3.7,4.2\right),\left(4.7,5.2\right),(5.6,6.28)$

$f\left(x\right)=2\cos \left(12\left(x+\frac{\pi }{4}\right)\right)+3$

This graph is periodic with a period of $2\pi .$

$\frac{\pi }{3}$

$\frac{\pi }{2}$

$\sqrt{1-{\left(1-2x\right)}^2}$

$\frac{1}{\sqrt{1+x^4}}$

$\frac{x+1}{x}$

False

approximately 0.07 radians