Advanced Functions 12
Advanced Functions 12
1st Edition
Chris Kirkpatrick, Kristina Farentino, Susanne Trew
ISBN: 9780176678326
Textbook solutions

All Solutions

Page 314: Getting Started

Exercise 1
Step 1
1 of 2
#### (a)

$$
28^circ
$$

#### (b)

$3360^circ-28^circ=332^circ$

Result
2 of 2
(a) $28^circ$; (b) $332^circ$
Exercise 2
Step 1
1 of 3
#### (a)

Side opposite: $-4$

Side adjacent: $3$

Hypotenuse: $h^2=3^2+4^2$

$h^2=25$

$h=5$

$sin theta=-dfrac{4}{5}$, $cos theta=dfrac{3}{5}$, $tan theta=-dfrac{4}{3}$,

$csctheta=-dfrac{5}{4}$, $sectheta=dfrac{5}{3}$, $cottheta=-dfrac{3}{4}$

Exercise scan

Step 2
2 of 3
#### (b)

$theta=sin^{-1}left(-dfrac{4}{5} right)$

$theta=307^{circ}$

The principal angle is $360^{circ}-53^{circ}=307^{circ}$

Result
3 of 3
(a) see solution; (b) $307^circ$.
Exercise 3
Step 1
1 of 7
#### (a)

$sin60^{circ}=dfrac{sqrt{3}}{2}$

Exercise scan

Step 2
2 of 7
#### (b)

$tan 180^{circ}=dfrac{0}{1}=0$

Exercise scan

Step 3
3 of 7
#### (c)

$sin 120^{circ}=dfrac{sqrt{3}}{2}$

Exercise scan

Step 4
4 of 7
#### (d)

$cos 300^{circ}=dfrac{1}{2}$

Exercise scan

Step 5
5 of 7
#### (e)

$sec 135^{circ}=-dfrac{sqrt{2}}{1}=-sqrt{2}$

Exercise scan

Step 6
6 of 7
#### (f)

$csc 270^{circ}=dfrac{1}{-1}=-1$

Exercise scan

Result
7 of 7
see solution
Exercise 4
Step 1
1 of 2
#### (a)

Since cosine is positive in the first and fourth quadrants, $theta=60^{circ}$,$300^{circ}$

#### (b)

Sincetangent is positive in the first and third quadrants, $theta=30^{circ}$,$210^{circ}$

#### (c)

Since tangent is positive in the first and third quadrants, $theta=45^{circ}$,$225^{circ}$

#### (d)

Cosine equals $-1$ at $theta=180^{circ}$

#### (e)

Contangent equals $-1$ at $theta=135^{circ}, 315^{circ}$

#### (f)

Sine equals $1$ at $theta=90^{circ}$

Result
2 of 2
see solution
Exercise 5
Step 1
1 of 3
#### (a)

$$
textbf{Period=$360^circ$, amplitude=$1$, $y=0, R=left{yinBbb{R}|-1 leq y leq 1 right}$}
$$

Exercise scan

Step 2
2 of 3
#### (b)

$$
textbf{Period=$360^circ$, amplitude=$1$, $y=0, R=left{yinBbb{R}| -1 leq y leq 1 right}$}
$$

Exercise scan

Result
3 of 3
see solution
Exercise 6
Step 1
1 of 3
#### (a)

$textbf{period}$ = $dfrac{2cdot180}{3}=120$, $y=0$, $45^circ$ to the left, $textbf{amplitude}$=$2$.

Exercise scan

Step 2
2 of 3
#### (a)

$textbf{period}$ = $dfrac{2cdot180}{dfrac{1}{2}}=720$, $y=-1$, $60^circ$ to the right, $textbf{amplitude}$=$1$.

Exercise scan

Result
3 of 3
see solution
Exercise 7
Step 1
1 of 2
$a$ is the amplitude, which determines how far above and below the axis of the curve of the function rises and falls; $k$ defines the period of the function, which is now often the function repeats itself; $d$ is the horizontal shift, which shifts the function to the right or the left; and $c$ is the vertical shift of the function.
Result
2 of 2
see solution
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