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I. FUNDAMENTALS
A. Real Numbers
(i) Graph the intervals (−5, 3] and (2, ∞) on the real number line.
Answer:
(ii) Express the inequalities x ≤ 3 and −1 ≤ x < 4 in interval notation.
Answer: (−∞, 3] and [−1, 4).
B. Exponents and Radicals
(i) Evaluate each expression.
(a)523
521(b)
(2
3
)−2
(c) 16−3/4
Solution: We have
(a)523
521= 523−21 = 52 = 25
(b)
(2
3
)−2
=
(3
2
)2
=22
32=
9
4
(c) 16−3/4 =1
163/4=
1
( 4√
16)3=
1
23=
1
8
(ii) Rationalize the denominator and simplify:
√10√
5 − 2
Solution: We have√
10√5 − 2
=
√10(
√5 + 2)
(√
5 − 2)(√
5 + 2)=
√10(
√5 + 2)
(√
5)2 − 22
=
√10(
√5 + 2)
5 − 4
=
√10(
√5 + 2)
1
=√
10(√
5 + 2)
=√
10√
5 + 2√
10
=√
2 · 5√
5 + 2√
10
=√
2√
5√
5 + 2√
10 = 5√
2 + 2√
10
1
C. Algebraic Expressions
(i) Simplify each expression. Write your final answer without negative exponents.
(a) (3a3b3)(4ab2)2 (b)
(3x3/2y3
x2y−1/2
)−2
(c)x2
x2 − 4− x + 1
x + 2(d)
y
x− x
y1
y− 1
x
Solution: We have
(a) (3a3b3)(4ab2)2 = (3a3b3)(42a2(b2)2) = (3a3b3)(16a2b4) = 48a3+2b3+4 = 48a5b7
(b)
(3x3/2y3
x2y−1/2
)−2
=
(x2y−1/2
3x3/2y3
)2
=
(x2−3/2
3y3+1/2
)2
=
(x1/2
3y7/2
)2
=(x1/2)2
32(y7/2)2=
x
9y7
(c)x2
x2 − 4− x + 1
x + 2=
x2
(x − 2)(x + 2)− x + 1
x + 2=
x2
(x − 2)(x + 2)− (x − 2)(x + 1)
(x − 2)(x + 2)
=x2 − (x − 2)(x + 1)
(x − 2)(x + 2)
=x2 − (x2 − x − 2)
(x − 2)(x + 2)
=x2 − x2 + x + 2
(x − 2)(x + 2)
=x + 2
(x − 2)(x + 2)
=1
x − 2
(d)
y
x− x
y1
y− 1
x
=
xy
(y
x− x
y
)
xy
(1
y− 1
x
) =
xy · y
x− xy · x
y
xy · 1
y− xy · 1
x
=
xy2
x− x2y
yxy
y− xy
x
=y2 − x2
x − y
=(y − x)(y + x)
x − y
= −(x − y)(y + x)
x − y
= −y + x
1
= −(y + x)
= −y − x
2
(ii) Perform the indicated operations and simplify.
(a) 3(x + 6) + 4(2x − 5) (b) (x + 3)(4x − 5) (c) (2x + 3)2 (d) (x + 2)3
Solution:
(a) We have
3(x + 6) + 4(2x − 5) = 3 · x + 3 · 6 + 4 · 2x − 4 · 5 = 3x + 18 + 8x − 20 = 11x − 2
(b) We have
(x + 3)(4x − 5) = x · 4x − x · 5 + 3 · 4x − 3 · 5 = 4x2 − 5x + 12x − 15 = 4x2 + 7x − 15
(c) We have(2x + 3)2 = (2x)2 + 2 · 2x · 3 + 32 = 4x2 + 12x + 9
since (a + b)2 = a2 + 2ab + b2.
(d) We have(x + 2)3 = x3 + 3x2 · 2 + 3x · 22 + 23 = x3 + 6x2 + 12x + 8
since (a + b)3 = a3 + 3a2b + 3ab2 + b3.
(iii) Factor each expression completely.
(a) 2x2 + 5x − 12 (b) x3 − 3x2 − 4x + 12 (c) x4 + 27x (d) x3y − 4xy
Solution:
(a) We first find two numbers a and b such that
a + b = 5 and ab = 2(−12) = −24
One can see that a = −3 and b = 8. We have
2x2 + 5x − 12 = 2x2 − 3x + 8x − 12 = x(2x − 3) + 4(2x − 3) = (2x − 3)(x + 4)
(b) We have
x3 − 3x2 − 4x + 12 = x2(x − 3) − 4(x − 3) = (x − 3)(x2 − 4) = (x − 3)(x2 − 22)
= (x − 3)(x − 2)(x + 2)
since a2 − b2 = (a − b)(a + b).
(c) We havex4 + 27x = x(x3 + 27) = x(x3 + 33) = x(x + 3)(x2 − 3x + 9)
since a3 + b3 = (a + b)(a2 − ab + b2).
(d) We havex3y − 4xy = xy(x2 − 4) = xy(x2 − 22) = xy(x − 2)(x + 2)
since a2 − b2 = (a − b)(a + b).
3
D. Equations
Find all real solutions.
(a) x+5 = 14− 1
2x (b)
2x
x + 1=
2x − 1
x(c) x2−x−12 = 0 (d) 2x2+4x+1 = 0
(e)√
3 −√
x + 5 = 2 (f) x4 − 3x2 + 2 = 0 (g) 3|x − 4| = 10
Solution:
(a) We have
x + 5 = 14 − 1
2x
2(x + 5) = 2
(
14 − 1
2x
)
2x + 10 = 28 − x
2x + 10 + x = 28 − x + x
3x + 10 = 28
3x + 10 − 10 = 28 − 10
3x = 18
3x
3=
18
3
x = 6
(b) We have2x
x + 1=
2x − 1
x
x(x + 1)2x
x + 1= x(x + 1)
2x − 1
x
2x2(x + 1)
x + 1=
(2x − 1)x(x + 1)
x
2x2 = (2x − 1)(x + 1)
2x2 = 2x2 + 2x − x − 1
2x2 = 2x2 + x − 1
2x2 − 2x2 = 2x2 + x − 1 − 2x2
0 = x − 1
x = 1
4
(c) Solution 1: We havex2 − x − 12 = 0
(x + 3)(x − 4) = 0
x = −3 or x = 4
Solution 2: We have
x1,2 =−(−1) ±
√
(−1)2 − 4 · 1 · (−12)
2 · 1 =1 ±
√49
2=
1 ± 7
2
therefore
x =1 − 7
2= −3 or x =
1 + 7
2= 4
(d) We have
x1,2 =−4 ±
√42 − 4 · 2 · 12 · 2 =
−4 ±√
8
4=
−4 ± 2√
2
4= −1 ±
√2
2
therefore
x = −1 −√
2
2or x = −1 +
√2
2
(e) We have√
3 −√
x + 5 = 2
(√
3 −√
x + 5)2 = 22
3 −√
x + 5 = 4√
x + 5 = −1
Since square root can not be negative, it follows that this equation has no solutions.
(f) If we set u = x2, then we get a quadratic equation in the new u:
u2 − 3u + 2 = 0
(u − 1)(u − 2) = 0
u = 1 or u = 2
x2 = 1 or x2 = 2
x = ±1 or x = ±√
2
(g) We have3|x − 4| = 10
|x − 4| =10
3
x − 4 = −10
3or x − 4 =
10
3
x =2
3or x =
22
3
5
E. Modeling with Equations
(i) Mary drove from Amity to Belleville at a speed of 50 mi/h. On the way back, she drove at
60 mi/h. The total trip took 42
5h of driving time. Find the distance between these two cities.
Solution: Let d represent the distance between the two cities. Let t represent the amount oftime she drove 50 mph. Then since 4 hours 24 minutes equals 4.4 hours, 4.4− t represents theamount of time driven at 60 mph.
Then d1 = 50t describes the outbound trip and d2 = 60(4.4 − t) describes the inbound trip,and, unless somebody picked Amity up and moved it while she was on the road or in Belleville,d1 = d2. Therefore:
50t = 60(4.4 − t) =⇒ 50t = 264 − 60t =⇒ 110t = 264 =⇒ t =264
110=
12
5
Therefore d = 50t = 50 · 12
5= 120 mi.
(ii) A rectangular parcel of land is 70 ft longer than it is wide. Each diagonal between oppositecomers is 130 ft. What are the dimensions of the parcel?
Solution: Let w = width, then w + 70 = length. Applying Pythagorean theorem:
w2 + (w + 70)2 = 1302
w2 + w2 + 2 · 70w + 702 = 16900
2w2 + 140w + 4900 = 16900
2w2 + 140w − 12000 = 0
w2 + 70w − 6000 = 0
(w − 50)(w + 120) = 0
w = 50 or w = −120
Toss out the negative solution: width is 50 feet, length is 50+70= 120 feet.
(iii) A bottle of medicine is to be stored at a temperature between 5◦C and 10◦C. Whatrange does this correspond to on the Fahrenheit scale? [Note: Fahrenheit (F ) and Celsius (C)
temperatures satisfy the relation C =5
9(F − 32).]
Solution: We have
5 ≤ 5
9F − 5
9· 32 ≤ 10
5 ≤ 5
9F − 160
9≤ 10
5 +160
9≤ 5
9F ≤ 10 +
160
9
41 ≤ F ≤ 50
6
F. Inequalities
Solve each inequality. Write the answer using interval notation, and sketch the solution on thereal number line.
(a) −4 < 5− 3x ≤ 17 (b) x(x− 1)(x + 2) > 0 (c) |x− 4| < 3 (d)2x − 3
x + 1≤ 1
Solution:
(a) We have−4 < 5 − 3x ≤ 17
−4 − 5 < −3x ≤ 17 − 5
−9 < −3x ≤ 12
12
−3≤ x <
−9
−3
−4 ≤ x < 3
(b) We know that the corresponding equation x(x−1)(x+2) = 0 has the solutions 0, 1, and −2.As shown in the Figure below, the numbers 0, 1, and −2 divide the real line into four intervals:(−∞,−2), (−2, 0), (0, 1), and (1,∞).
-• • •-2 0 1
On each of these intervals we determine the signs of the factors using test values. We choosea number inside each interval and check the sign of the factors x, x− 1, and x + 2 at the valueselected. For instance, if we use the test value x = −3 for the interval (−∞,−2) shown inFigure above, then substitution in the factors x, x − 1, and x + 2 gives
x = −3, x − 1 = −3 − 1 = −4, x + 2 = −3 + 2 = −1
All three factors are negative on this interval, therefore x(x−1)(x+2) is negative on (−∞,−2).Similarly, using the test values x = −1, x = 1/2 and x = 2 for the intervals (−2, 0), (0, 1) and(1,∞), respectively, we get:
-• • •-2 0 1– + – +
Thus, the solution of the inequality x(x − 1)(x + 2) > 0 is (−2, 0) ∪ (1,∞).
7
(c) We have
|x − 4| < 3 =⇒ −3 < x − 4 < 3 =⇒ −3 + 4 < x < 3 + 4 =⇒ 1 < x < 7
(d) We have2x − 3
x + 1≤ 1
2x − 3
x + 1− 1 ≤ 0
2x − 3
x + 1− x + 1
x + 1≤ 0
2x − 3 − (x + 1)
x + 1≤ 0
2x − 3 − x − 1
x + 1≤ 0
x − 4
x + 1≤ 0
As shown in the Figure below, the numbers 4 (at which the numerator ofx − 4
x + 1is 0) and −1 (at
which the denominator ofx − 4
x + 1is 0) divide the real line into three intervals: (−∞,−1), (−1, 4),
and (4,∞).
-• •-1 4
On each of these intervals we determine the sign ofx − 4
x + 1using test values. We choose a
number inside each interval and check the sign ofx − 4
x + 1at the value selected. For instance, if
we use the test value x = −2 for the interval (−∞,−1) shown in Figure above, then substitution
inx − 4
x + 1gives
−2 − 4
−2 + 1=
−6
−1= 6 > 0
Similarly, using the test values x = 0 and x = 5 for the intervals (−1, 4), and (4,∞), respectively,we get:
-• •-1 4+ – +
Thus, the solution of the inequality2x − 3
x + 1≤ 1 is (−1, 4].
8
G. Coordinate Geometry.
(i) (a) Plot the points P (0, 3), Q(3, 0), and R(6, 3) in the coordinate plane. Where must thepoint S be located so that PQRS is a square?
Answer: S(3, 6)
(b) Find the area of PQRS.
Solution: Let d = |PQ|. Since |SQ| = 6, we have
d2 + d2 = 62 =⇒ 2d2 = 36 =⇒ d2 = 18
It follows that the area of PQRS is d2 = 18.
(ii) (a) Sketch the graph of y = x2 − 4.(b) Find the x-and y-intercepts of the graph.(c) Is the graph symmetric about the x-axis, the y-axis, or the origin?
Solution:
(a)
(b) To find the x-intercepts, we solvex2 − 4 = 0
It follows that the x-intercepts are −2 and 2. To find the y-intercept, we set x = 0 in y = x2−4.It follows that the y-intercept is y = 02 − 4 = −4.(c) The graph is symmetric about the y-axis only.
9
(iii) Find the center and radius of each circle and sketch its graph.
(a) x2 + y2 = 25 (b) (x − 2)2 + (y + 1)2 = 9 (c) x2 + 6x + y2 − 2y + 6 = 0
Solution:
(a) The center is at (0, 0) and the radius is 5.
(b) The center is at (2,−1) and the radius is 3.
(c) We have
x2 + 6x + y2 − 2y + 6 = (x2 + 6x + 9) + (y2 − 2y + 1) − 4 = (x + 3)2 + (y − 1)2 − 22 = 0
Therefore the center is at (−3, 1) and the radius is 2.
10
H. Lines
(i) Let P (−3, 1) and Q(5, 6) be two points in the coordinate plane.(a) Plot P and Q in the coordinate plane.(b) Find the distance between P and Q.(c) Find the midpoint of the segment PQ.(d) Find the slope of the line that contains P and Q.(e) Find the perpendicular bisector of the line that contains P and Q.(f) Find an equation for the circle for which the segment PQ is a diameter.
Solution:
(a)
(b) The distance between P and Q is
d =√
(−3 − 5)2 + (1 − 6)2 =√
(−8)2 + (−5)2 =√
64 + 25 =√
89
(c) The midpoint of the segment PQ is(−3 + 5
2,1 + 6
2
)
=
(2
2,7
2
)
=
(
1,7
2
)
(d) The slope of the line that contains P and Q is
m =1 − 6
−3 − 5=
−5
−8=
5
8
(e) It follows from (d) that the slope of the perpendicular bisector of the line that contains P
and Q is −8
5, therefore an equation of the perpendicular bisector of the line that contains P
and Q is
y − 7
2= −8
5(x − 1)
which can be rewritten as
y = −8
5x +
51
10
(f) Since the midpoint of the segment PQ is
(
1,7
2
)
and the distance between P and Q is√
89,
it follows that an equation for the circle for which the segment PQ is a diameter is
(x − 1)2 +
(
y − 7
2
)2
=89
4
11
(ii) Write the linear equation 2x− 3y = 15 in slope-intercept form, and sketch its graph. Whatare the slope and y-intercept?Solution: We have
2x − 3y = 15 =⇒ −3y = −2x + 15 =⇒ y =2
3x − 5
Therefore the slope is2
3and y-intercept is −5.
(iii) Find an equation for the line with the given property.(a) It passes through the point (3,−6) and is parallel to the line 3x + y − 10 = 0.(b) It has x-intercept 6 and y-intercept 4.
Solution:
(a) We have3x + y − 10 = 0 =⇒ y = −3x + 10
Therefore the slope of the line is −3. It follows that an equation for the line that passes throughthe point (3,−6) and is parallel to the line 3x + y − 10 = 0 is
y − (−6) = −3(x − 3) =⇒ y = −3x + 3
(b) Since the y-intercept is 4, it follows that b = 4. Since the x-intercept is 6, it follows that
0 = 6m + b
Plugging in 4 into this equation, we get m = −2
3. Therefore an equation for the line that has
x-intercept 6 and y-intercept 4 is
y = −2
3x + 4
12
II. FUNCTIONS
A. Definition and Graphs of Functions
(i) Which of the following are graphs of functions? If the graph is that of a function, is itone-to-one?
Answer: (a) and (b) are graphs of functions, (a) is one-to-one.
(ii) Let f(x) =
√x + 1
x.
(a) Evaluate f(3), f(5), and f(a − 1).(b) Find the domain of f .
Solution:
(a) We have
f(3) =
√3 + 1
3=
√4
3=
2
3
f(5) =
√5 + 1
5=
√6
5
and
f(a − 1) =
√a − 1 + 1
a − 1=
√a
a − 1
(b) The domain of f is [−1, 0) ∪ (0,∞).
13
B. Average Rate of Change
Determine the average rate of change for the function f(t) = t2 − 2t between t = 2 and t = 5.
Solution: Since the average rate of change of a function f between two points t = a and t = bis
f(b) − f(a)
b − a
we have
Average rate =(52 − 2 · 5) − (22 − 2 · 2)
5 − 2=
15 − 0
3= 5
C. Transformations of Functions
(i) How is the graph of y = f(x − 3) + 2 obtained from the graph of f?
Answer: The graph shifts right 3 units, then shifts upward 2 units.
(ii) How is the graph of y = f(−x) obtained from the graph of f?
Answer: The graph reflects about the y-axis.
D. Quadratic Functions; Maxima and Minima
(i) (a) Write the quadratic function f(x) = 2x2 − 8x + 13 in standard form.(b) Sketch a graph of f .(c) What is the minimum value of f?
Solution:
(a) We have
f(x) = 2x2 − 8x + 13 = 2x2 − 8x + 8 + 5 = 2(x2 − 4x + 4) + 5 = 2(x − 2)2 + 5
(b)
(c) Since (x − 2)2 ≥ 0, it follows that
f(x) = 2(x − 2)2 + 5 ≥ 5
Therefore the minimum value of f is 5.
14
(ii) Let f(x) =
{
1 − x2 if x ≤ 0
2x + 1 if x > 0
(a) Evaluate f(−2) and f(1).(b) Sketch the graph of f .
Solution:
(a) We havef(−2) = 1 − (−2)2 = 1 − 4 = −3, f(1) = 2 · 1 + 1 = 3
(b)
E. Modeling with Functions
If 1800 ft of fencing is available to build five adjacent pens, as shown in the diagram below,express the total area of the pens as a function of x. What value of x will maximize the totalarea?
Solution: The function that models the area will be
A(x) = −3x2 + 900x
where x is the width of the pens. It follows that the area will attain its maximum when
x =900
2 · 3 = 150 ft
Using this we can find that the maximum area will be
A(x) = −3 · 1502 + 900 · 150 = 67500 ft
15
F. Combining Functions
If f(x) = x2 + 1 and g(x) = x − 3, find the following.
(a) f ◦ g (b) g ◦ f (c) f(g(2)) (d) g(f(2)) (e) g ◦ g ◦ g
Solution: We have
(a) f ◦ g = (x − 3)2 + 1
(b) g ◦ f = x2 + 1 − 3 = x2 − 2
(c) f(g(2)) = (2 − 3)2 + 1 = 2
(d) g(f(2)) = 22 − 2 = 2
(e) g ◦ g ◦ g = x − 3 − 3 − 3 = x − 9
G. One-to-One Functions and Their Inverses
(i) (a) If f(x) =√
3 − x, find the inverse function f−1.(b) Sketch the graphs of f and f−1 on the same coordinate axes.
Solution:
(a) We have:
Step 1: Replace f(x) by y:y =
√3 − x
Step 2: Solve for x:
y =√
3 − x =⇒ y2 = 3 − x =⇒ x = 3 − y2
Step 3: Replace x by f−1(x) and y by x:
f−1(x) = 3 − x2
Since the range of f(x) is all nonnegative numbers, it follows that the domain of f−1(x) isx ≥ 0. So,
f−1(x) = 3 − x2, x ≥ 0
(b)
16
(ii) The graph of a function f is given.(a) Find the domain and range of f .(b) Sketch the graph of f−1.(c) Find the average rate of change of f between x = 2 and x = 6.
Answer:
(a) Domain [0, 6], range [1, 7].
(b)
(c)5
4
17
III. POLYNOMIAL AND RATIONAL FUNCTIONS
A. Polynomial Functions and Their Graphs
Graph the polynomial P (x) = −(x + 2)3 + 27, showing clearly all x- and y-intercepts.
Answer:
B. Dividing Polynomials
Use long division to find the quotient and remainder when 2x5 + 4x4 − x3 − x2 + 7 is dividedby 2x2 − 1.
Answer: The quotient and remainder are x3 + 2x2 +1
2and
15
2, respectively.
C. Rational Functions
Consider the following rational functions:
r(x) =2x − 1
x2 − x − 2s(x) =
x3 + 27
x2 + 4t(x) =
x3 − 9x
x + 2u(x) =
x2 + x − 6
x2 − 25
(a) Which of these rational functions has a horizontal asymptote?(b) Which of these functions has a slant asymptote?(c) Which of these functions has no vertical asymptote?(d) Graph y = u(x), showing clearly any asymptotes and x-and y-intercepts the function
Answer:
(a) Only r and u have horizontal asymptotes.
(b) Only s has a slant asymptote.
(c) Only s has no vertical asymptote.
(d)
18
IV. EXPONENTIAL AND LOGARITHMICFUNCTIONS
A. Exponential Functions
Graph the function y = 2x−3.
Answer:
B. Logarithmic Functions
Sketch the graph of the function f(x) = log(x+1) and state the domain, range, and asymptote.
Answer: Domain (−1,∞), range (−∞,∞), asymptote x = −1.
C. Laws of Logarithms
(i) Evaluate each logarithmic expression.
(a) log3
√27 (b) log2 80 − log2 10 (c) log8 4
Solution: We have
(a) log3
√27 = log3
√33 = log3 33/2 =
3
2log3 3 =
3
2· 1 =
3
2
(b) log2 80 − log2 10 = log2
80
10= log2 8 = log2 23 = 3 log2 2 = 3 · 1 = 3
(c) log8 4 = log8 22 = log8(3√
8)2 = log8 82/3 =2
3log8 8 =
2
3· 1 =
2
3
19
(ii) Expand: log 3
√x + 2
x4(x2 + 4).
Solution: We have
log 3
√
x + 2
x4(x2 + 4)= log
(x + 2)1/3
x4/3(x2 + 4)1/3=
1
3log(x + 2) − 4
3log x − 1
3log(x2 + 4)
(iii) Combine into a single logarithm: ln x − 2 ln(x2 + 1) +1
2ln(3 − x4).
Solution: We have
ln x − 2 ln(x2 + 1) +1
2ln(3 − x4) = ln x − ln(x2 + 1)2 + ln
√3 − x4 = ln
x√
3 − x4
(x2 + 1)2
D. Exponential and Logarithmic Equations
Find the solution of the equation, correct to two decimal places.
(a) 10x+3 = 62x (b) 5 ln(3 − x) = 4 (c) log2(x + 2) + log2(x − 1) = 2
Solution:
(a) We have10x+3 = 62x
ln 10x+3 = ln 62x
(x + 3) ln 10 = 2x ln 6
x ln 10 + 3 ln 10 = 2x ln 6
x ln 10 − 2x ln 6 = −3 ln 10
x(ln 10 − 2 ln 6) = −3 ln 10
x =−3 ln 10
ln 10 − 2 ln 6≈ 5.39
(b) We have5 ln(3 − x) = 4
ln(3 − x) =4
5
3 − x = e4/5
x = 3 − e4/5 ≈ 0.77
(c) We havelog2(x + 2) + log2(x − 1) = 2
log2(x + 2)(x − 1) = log2 4
(x + 2)(x − 1) = 4
x = 2 or x = −3
Since x = −3 is not from the domain of log2(x + 2) + log2(x − 1), the only answer is x = 2.
20
E. Modeling with Exponential and Logarithmic Functions
(i) The initial size of a culture of bacteria is 1000. After one hour the bacteria count is 8000.(a) Find a function that models the population after t hours.(b) Find the population after 1.5 hours.(c) When will the population reach 15,000?(d) Sketch the graph of the population function.
Answer: (a) n(t) = 1000e2.07944t (b) 22, 627 (c) 1.3 h (d)
(ii) Suppose that $12,000 is invested in a savings account paying 5.6% interest per year.(a) Write the formula for the amount in the account after t years if interest is compounded
monthly.(b) Find the amount in the account after 3 years if interest is compounded daily.(c) How long will it take for the amount in the account to grow to $20,000 if interest is
compounded semiannually?
Answer: (a) A(t) = 12, 000
(
1 +0.056
12
)12t
(b) $14, 195.06 (c) 9.249 yr
21
V. TRIGONOMETRIC FUNCTIONS OF REALNUMBERS
A. The Unit Circle
The point P (x, y) is on the unit circle in quadrant IV. If x =√
11/6, find y.
Solution: We have
y = −√
1 − x2 = −
√√√√1 −
(√11
6
)2
= −√
1 − 11
36= −
√
25
36= −5
6
B. Trigonometric Functions of Real Numbers
(i) The point P in the figure at the right has y-coordinate4
5. Find:
(a) sin t (b) cos t (c) tan t (d) sec t
Answer: (a)4
5(b) −3
5(c) −4
3(d) −5
3
(ii) Find the exact value.
(a) sin7π
6(b) cos
13π
4(c) tan
(
−5π
3
)
(d) csc3π
2
Answer: (a) −1
2(b) −
√2
2(c)
√3 (d) −1
(iii) Express tan t in terms of sin t, if the terminal point determined by t is in quadrant II.
Solution: We have
tan t =sin t
cos t=
sin t
±√
1 − sin2 t= [since the terminal point is in quadrant II] = − sin t
√
1 − sin2 t
(iv) If cos t = − 8
17and if the terminal point determined by t is in quadrant III, find
tan t cot t + csc t
Solution: We have
tan t cot t + csc t = 1 +1
sin t= 1 +
1
±√
1 − cos2 t
Therefore if cos t = − 8
17and if the terminal point determined by t is in quadrant III, we get
tan t cot t + csc t = 1 − 1√1 − cos2 t
= 1 − 1√
1 −(
− 8
17
)2= 1 − 17
15= − 2
15
22
C. Trigonometric Graphs
(i) Find the amplitude, period, and phase shift of the function y = 2 sin
(1
2x − π
6
)
. Sketch
the graph.
Answer: Amplitude 2, period 4π, phase shift π/3.
(ii) The graph shown below is one period of a function of the form y = a sin k(x−b). Determinethe function.
Answer: y = 2 sin 2(x + π/3)
23
VI. TRIGONOMETRIC FUNCTIONS OF ANGLES
A. Angle Measure
(i) Find the radian measures that correspond to the degree measures 330◦ and -135◦.
Answer: 11π/6 and −3π/4
(ii) Find the degree measures that correspond to the radian measures4π
3and −1.3.
Answer: 240◦ and −74.5◦
B. Trigonometry of Right Triangles
Find tan θ + sin θ for the angle θ shown.
Solution: Since the hypotenuse of the triangle is√
22 + 32 =√
13, it follows that
tan θ + sin θ =2
3+
2√13
=2√
13 + 6
3√
13=
26 + 6√
13
39
C. Trigonometric Functions of Angles
Find the exact value of each of the following.
(a) sin 405◦ (b) tan(−150◦) (c) sec5π
3(d) csc
5π
2
Answer: (a)√
2/2 (b)√
3/3 (c) 2 (d) 1
24
D. The Law of Sines
Refer to the figure below. Find the side labeled x.
Solution: We havex
sin 69◦=
230
sin(180◦ − 52◦ − 69◦︸ ︷︷ ︸
59◦
)
therefore
x =230 sin 69◦
sin 59◦≈ 250.5
E. The Law of Cosines
Refer to the figure below.(a) Find the angle opposite the longest side.(b) Find the area of the triangle.
Solution:
(a) By the Law of Cosines we have
202 = 132 + 92 − 2 · 13 · 9 cosα
therefore
cos α =132 + 92 − 202
2 · 13 · 9 = −25
39It follows that
α = arccos
(
−25
39
)
≈ 129.9◦
(b) The semiperimeter of the triangle is
s =1
2(9 + 13 + 20) = 21
therefore by Heron’s formula the area of the triangle is
A =√
21(21 − 9)(21 − 13)(21 − 20) =√
21 · 12 · 8 · 1 =√
2016 = 12√
14 ≈ 44.9
25
VII. ANALYTIC TRIGONOMETRY
A. Trigonometric Identities
Verify each identity.
(a) tan θ sin θ + cos θ = sec θ (b)2 tanx
1 + tan2 x= sin 2x
Solution:
(a) We have
tan θ sin θ + cos θ =sin θ
cos θsin θ + cos θ =
sin2 θ
cos θ+ cos θ =
sin2 θ
cos θ+
cos2 θ
cos θ
=sin2 θ + cos2 θ
cos θ=
1
cos θ= sec θ
(b) We have
2 tanx
1 + tan2 x=
2sin x
cos x
1 +
(sin x
cos x
)2=
(
2sin x
cos x
)
cos2 x
(
1 +
(sin x
cos x
)2)
cos2 x
=2 sinx cos x
cos2 x + sin2 x=
2 sin x cos x
1= sin 2x
B. Addition and Subtraction Formulas
(i) Find the exact value of each expression.
(a) sin 8◦ cos 22◦ + cos 8◦ sin 22◦ (b) sin 75◦ (c) sinπ
12
Solution:
(a) We have
sin 8◦ cos 22◦ + cos 8◦ sin 22◦ = sin(8◦ + 22◦) = sin 30◦ =1
2
(b) We have
sin 75◦ = sin(45◦ + 30◦) = sin 45◦ cos 30◦ + cos 45◦ sin 30◦
=
√2
2·√
3
2+
√2
2· 1
2=
√6
4+
√2
4=
√6 +
√2
4
(c) We have
sinπ
12= sin
π/6
2=
√
1 − cos(π/6)
2=
√
1 −√
3/2
2=
√
2 −√
3
4=
1
2
√
2 −√
3
26
(ii) For the angles α and β in the figures, find cos(α + β).
Solution: We have
cos(α + β) = cos α cos β − sin α sin β =2√5·√
5
3− 1√
5· 2
3=
2√
5 − 2
3√
5=
10 − 2√
5
15
C. Double-Angle, Half-Angle, and Sum-Product Identities
(i) (a) Write sin 3x cos 5x as a sum of trigonometric functions.(b) Write sin 2x − sin 5x as a product of trigonometric functions.
Solution:
(a) We have
sin 3x cos 5x =1
2[sin(3x + 5x) + sin(3x − 5x)] =
1
2[sin(8x) + sin(−2x)] =
1
2[sin(8x) − sin(2x)]
(b) We have
sin 2x − sin 5x = 2 cos2x + 5x
2sin
2x − 5x
2= 2 cos
7x
2sin
−3x
2= −2 cos
7x
2sin
3x
2
(ii) If sin θ = −4
5and θ is in quadrant III, find tan(θ/2).
Solution: We have
tan(θ/2) =1 − cos θ
sin θ=
1 +√
1 − sin2 θ
sin θ=
1 +
√
1 −(
−4
5
)2
−4
5
=1 +
3
5
−4
5
= −2
27
D. Inverse Trigonometric Functions
(i) Graph y = sin x and y = sin−1 x, and specify the domain of each function.
Answer:
(ii) Find the exact value of cos
(
tan−19
40
)
.
Solution 1: We have
cos x = ± 1√1 + tan2 x
Since −π/2 < tan−1 θ < π/2, it follows that cos (tan−1 θ) > 0. Therefore
cos x =1√
1 + tan2 x
hence
cos
(
tan−19
40
)
=1
√
1 + tan2
(
tan−19
40
) =1
√
1 +
(9
40
)2
=40√
402 + 92
=40
41
Solution 2: Put θ = tan−19
40, so tan θ =
9
40. Then
cos
(
tan−19
40
)
= cos θ =40
41
����������
θ
9
40
41
28
E. Trigonometric Equations
Solve each trigonometric equation in the interval [0, 2π).
(a) 2 cos2 x + 5 cosx + 2 = 0 (b) sin 2x − cos x = 0
Solution:
(a) We have2 cos2 x + 5 cos x + 2 = (2 cosx + 1)(cosx + 2) = 0
therefore2 cos x + 1 = 0 or cosx + 2 = 0
Since the second equation is impossible, it follows that
cos x = −1
2
which gives x =2π
3,
4π
3.
(b) We havesin 2x − cos x = 2 sinx cos x − cos x = cos x(2 sin x − 1) = 0
therefore
cos x = 0 or sin x =1
2
which gives x =π
6,
π
2,
5π
6,
3π
2.
29
VIII. POLAR COORDINATES AND VECTORS
A. Polar Coordinates
(i) Convert the point whose polar coordinates are (8, 5π/4) to rectangular coordinates.
Solution: We have
x = 8 cos5π
4= 8
(
−√
2
2
)
= −4√
2, y = 8 sin5π
4= 8
(
−√
2
2
)
= −4√
2
(ii) Find two polar coordinate representations for the rectangular coordinate point (−6, 2√
3),one with r > 0 and one with r < 0, and both with 0 ≤ θ < 2π.
Solution: We have
r = ±√
(−6)2 + (2√
3)2 = ±√
48 = ±4√
3
and
tan θ =2√
3
−6= − 1√
3
It follows that two polar coordinate representations are (4√
3, 5π/6) and (−4√
3, 11π/6).
B. Graphs of Polar Equations
Graph the polar equation r = 8 cos θ. What type of curve is this? Convert the equation torectangular coordinates.
Solution: We multiply both sides of the equation by r, because then we can use the formulasr2 = x2 + y2 and r cos θ = x. We have
r = 8 cos θ
r2 = 8r cos θ
x2 + y2 = 8x
x2 − 8x + 16 + y2 = 16
(x − 4)2 + y2 = 42
which is a circle of radius 4 and center (4, 0).
30
C. Vectors
Let u be the vector with initial point P (3,−1) and terminal point Q(−3, 9).(a) Express u in terms of i and j.(b) Find the length of u.
Solution:
(a) u = (−3 − 3)i + (9 − (−1))j = −6i + 10j
(b) |u| =√
(−6)2 + 102 =√
136 = 2√
34
D. The Dot Product
(i) Let u = 〈1, 3〉 and v = 〈−6, 2〉.(a) Find u− 3v.(b) Find |u + v|.(c) Find u · v.(d) Are u and v perpendicular?
Solution:
(a) We haveu− 3v = 〈1 − 3(−6), 3 − 3 · 2〉 = 〈19,−3〉
(b) Since u + v = 〈1 + (−6), 3 + 2〉 = 〈−5, 5〉, it follows that
|u + v| =√
(−5)2 + 52 = 5√
2
(c) We haveu · v = 1 · (−6) + 3 · 2 = 0
(d) Since u · v = 0, it follows that u and v are perpendicular.
(ii) Let u = 3i + 2j and v = 5i − j.(a) Find the angle between u and v.(b) Find the component of u along v.(c) Find proj
vu.
Solution: We first find |u|, |v| and u · v:
|u| =√
32 + 22 =√
13, |v| =√
52 + (−1)2 =√
26, u · v = 3 · 5 + 2 · (−1) = 13
(a) We have
cos θ =u · v|u||v| =
13√13√
26=
1√2
therefore the angle between u and v is π/4.
(b) The component of u along v is |u| cos θ =√
13 · 1√2
=
√26
2.
(c) We have
projvu =
(u · v|v|2
)
v =
(13
(√
26)2
)
(5i − j) =5
2i − 1
2j
31
IX. SYSTEMS OF EQUATIONS AND INEQUALITIES
A. Systems of Linear Equations in Two Variables
Find all solutions of the system.
(a)
{x + 3y = 7
5x + 2y = −4(b)
{6x + y2 = 10
3x − y = 5
Answer: (a) x = −2, y = 3. (b) x = 1, y = −2 or x = 5/3, y = 0.
B. Systems of Linear Equations in Several Variables
Find all solutions of the system or show that no solution exists.
(a)
x − y + 2z = 0
2x − 4y + 5z = −5
2y − 3z = 5 = −2
(b)
2x − 3y + z = 3
x + 2y + 2z = −1
4x + y + 5z = 4
Answer: (a) x = 5/2, y = 5/2, z = 0. (b) No solution.
C. Systems of Inequalities
Graph the solution set of the system of inequalities. Label the vertices with their coordinates.
(a)
2x + y ≤ 8
x − y ≥ −2
x + 2y ≥ 4
(b)
{x2 + y ≤ 5
y ≤ 2x + 5
Answer:
32