MTH 111 Exam III Practice Problems Answers/Hints
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Note: Calculators WILL be permitted, in fact required, for this exam.
Formulas will NOT be provided. You should know the exact expressions
for the sine and cosine of π/6, π/4, π/3, π/2 and π and be able to deduce the
sines, cosines and tangents, etc. of related angles from these. You should
know the identities for sin(x±y) and cos(x±y), the identity sin2 x+cos2 x = 1
and related identities, and identities about periodicity, e.g. cos(x + 2π) =
cos(x), cos(x + π) = − cos(x) and similar identities for sine, and the identity
cos(x − π/2) = sin(x). These will not be provided.
The exam will have only one part, with 10 questions like the following,
each worth 10 points. WORK MUST BE SHOWN FOR CREDIT. Partial
credit will be given when appropriate.
NOTE: Most of these problems can be done in several ways. The hints
and solutions below suggest one or two valid ways, but others are possible too.
3.1 a) A population of space gerbils has an exponential growth rate of 7.2%
per day. If the initial population is 100, find the population after 100 days has
passed.
b) At what time will the population reach 200? Solve this algebraically, not
by trial and error.
a) If P is the population at time t, P = 100e.072t , so the population at time
t = 100 is 100e.072·100 = 100e7.2 ≈ 133943.
(These are very acive gerbils.)
b) Solve the equation 200 = 100e.072t for t: 2 = e.072t =⇒ ln 2 = .072t =⇒
t = ln 2/.072 ≈ 9.6 (days)
3.2 a) If 102x = 1/1000 find x. (Give exact answer.)
b) If log2 (x) = −3, find x.
MTH 111 Exam III Practice Problems Answers/Hints
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a) This equation is equivalent to 2x = log10 (1/1000) or 2x = −3. Thus
x = −3/2.
b) This equation is equivalent to 2−3 = x so x = 2−3 = 1/23 = 1/8.
3.3 a) Find logb (1/b3 ).
b) Find the value of log2 (25) to 3 decimal places.
a) logb (1/b3 ) = −3 logb (b) = −3 · 1 = −3
b) log2 (25) =
ln 25
ln 2
≈ 4.649
3.4 Suppose you deposit $ 200 in an account which earns interest at 5.3% per
year compounded continuously. a) Write a formula for the value of your account
t years later. b) Determine when your account will be worth $ 500. c) Suppose
you need this account to be worth $ 500 within 10 years. Is the interest rate
sufficient? What is the minimum interest rate you would need to guarantee this
return on your $ 200 investment?
a) If A is the amount in your account at time t (years), A = 200e0.053t .
b) Solve 500 = 200e0.053t for t: 5/2 = e0.053t =⇒ ln(5/2) = .053t =⇒ t =
ln(2.5)/.053 ≈ 17.3
c) No. Solve 500 = 200er·10 for r to get the minimum interest rate needed.
r = ln(5/2)/10 ≈ .0916 so 9.2% would do.
3.5 The half-life of potassium-40 is 1.31 × 109 years. How long will it take 10
gms of potassium 40 to decay to 2 grams?
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1/2 = er×1.31×10 =⇒ r = ln(1/2)/(1.31 × 109 ) ≈ −5.29 × 10−10 . Solve
2 = 10ert for t. t = ln(.2)/r ≈ 3.04 × 109 so it will take about 3 billion years.
MTH 111 Exam III Practice Problems Answers/Hints
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3.6 a) Express as a single logarithm: log2 (x + 3) − 4 log2 (x − 3) + 4.
b) Solve the equation log3 (x2 − 1) = 2 algebraically. SHOW YOUR WORK!
a) log2 (16(x2 − 9))
√
b) x2 − 1 = 32 so x = ± 10
3.7 a) Find the period of the function f (x) = −2 cos( π3 x + 1).
b) Find the period of the function g(x) = 3 tan(x).
a) period is
2π
π/3
=6
b) period is π
3.8 a) A central angle α cuts out an arc of length 1/9 in a circle of radius 1.
Give the radian measure of this angle.
b) What is the radian measure (with correct sign) of the angle that the minute
hand of a clock moves through from 2:00 PM to 4:30 PM?
a) 1/9 = α · 1 so α = 1/9, or just recall the definition of radian measure for
angles.
b) The minute hand moves clockwise (of course) two and a half times
around the circle so the angle in radians that it moves through is −(2.5)(2π) =
−5π.
3.9 Find the exact value (no decimals, leave answer in terms of fractions and
radicals) the following angles using only knowledge of the sine and cosine of
π/2, π/3, and π/4 together with your understanding of radian measure. SHOW
YOUR WORK.
a) sin(−19π/3).
b) tan(9π/4)
MTH 111 Exam III Practice Problems Answers/Hints
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a) −19π/3 = −6π−π/3 = −3(2π)−π/3 so −19π/3 has the same terminal
side (and so the same sine) as −π/3 which whose sine is minus that of the
√
sine of π/3. Thus sin(−19π/3) = − sin(π/3) = − 3/2.
b) 9π/4 = 2π + π/4 so 9π/4 has the same terminal side as π/4 and so the
same tangent. Thus tan(9π/4) = tan(π/4) = 1.
3.10 a) Find the angle θ between 180o and 270o where
cos(θ) = −
√
3
θ, in radians, between 3π and 4π where sin(θ) = − 2 .
a) 210o
b) Either 3π +
π
3
=
10π
3
or 4π −
π
3
=
11π
3
√
3
.
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b) Find
(radians)
3.11 a) Express the range of the function f (x) = −3 cos(2x) + 2 in interval
notation.
b) What is the domain of the function g(x) = sin−1 (x)? Give your answer
using interval notation.
c) What is the range of the function f (x) = cos−1 (x)? Give your answer
using interval notation.
a) Since cos(2x) varies between −1 and 1, −3 cos(2x) will vary between −3
and 3 so −3 cos(2x) + 2 will vary between −1 and 5. The range is [−1, 5].
b) The domain of sin−1 (x) is [−1, 1].
c) The range of cos−1 (x) is [0, π].
3.12 Find
a) sin−1 (sin(π)).
b) sin(sin−1 (2/7)). (Give the exact answer only, no decimals, leave fractions
and radicals if needed).
MTH 111 Exam III Practice Problems Answers/Hints
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c) sin(cos−1 (−1/7)) exactly. (Give the exact answer only, no decimals, leave
fractions and radicals if needed).
a) sin−1 (0) = 0.
b) sin−1 (x) is the number in [−π/2, π2] whose sine is x, so sin(sin−1 (x)) =
x always. So sin(sin−1 (2/7)) = 2/7.
(Note that sin−1 (sin(x)) is NOT always equal to x.
c) If θ = cos−1 (−1/7) then cos(θ) = −1/7 and θ is between 0 and π.
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(This is the definition of the arc cosine.) So sin(θ)
q must be ≥√0 and sin (θ) =
1 − cos2 (θ) = 1 − 1/49 = 48/49 =⇒ sin(θ) = 48/49 (= 4 3/7).
3.13 a) If f (x) = 3 cos(x/2) for 0 ≤ x ≤ 2π, find a formula for f −1 (x).
b) If g(x) = e4x find a formula for g −1 (x)
a) f −1 (x) = 2 cos−1 (x/3)
b) g −1 (x) = 41 ln x
3.14 Suppose b is some real number that you are given. Let f (x) = tan( xb )
a) What is the exact value of the period of the function f (x)? (Your answer
should involve b.
b) Where does the graph of y = f (x) have vertical asymptotes?
c) Make a rought sketch of the graph of y = f (x) showing the asymptotes.
π
a) period is 1/b
= bπ
multiples of bπ/2
b) at all odd number (positive and negative)
3.15 Find sin(105o ) exactly. (Give the answer in terms of radicals and fractions, no decimals.) Hint: 105=45+60.
MTH 111 Exam III Practice Problems Answers/Hints
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√
√
Use the formula for sin(α + β). Answer: ( 6 + 2)/4
3.16 The angle φ is between 90o and 180o , and sin(φ) = 5/13. Find the exact
value of cos(φ) and tan(φ). Give the exact answer in terms of radicals and
fractions, no decimal.
First of all, cos(φ) must be negative. Then draw a triangle with hypotenuse in the second quadrant with length 13 and making angle φ with
the x-axis so that the side opposite φ has length 5. By the Pythagorean
theorem,
the side adjacent to φ, which lies along the x-axis, has length
q
2
(13 − 52 ) = 12, so cos(φ) = −12/13. Alternatively, use the identity
q
q
5 2
) =−
sin2 φ + cos2 φ = 1 to see that cos(φ) = − 1 − ( 13
Then tan(φ) = sin(φ)/ cos(φ) = −5/12.
144
169
= −12/13.
3.17 a) Describe how the graph of f (x) = cos(3x − π2 ) obtained from the
graph of f (x) = cos(3x) by shifting?
b) Describe how the graph of 3 sin(2x) can be obtained from the graph of
cos(2x) by shifting and stretching
a) The graph of y = cos(3x − π2 ) = cos(3(x − π6 )) is obtained from that of
y = cos(3x) by shifting π6 units to the right.
b) sin(2x) = cos(2x − π2 ) = cos(2(x − π4 )), so the graph of sin(2x) is
obtained from that of cos(2x) by shifting π/4 units to the right. The graph
of 3 sin(2x) is then obtained by stretching vertically by 3 units.
3.18 Write a formula for a SINGLE function of the form
f (x) = A sin(B(x − C)) + D with the following properties.
i) The period is π.
ii) f (x) varies between 2 and 10.
iii) f (0) = 8.
MTH 111 Exam III Practice Problems Answers/Hints
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From (i) and (ii) f (x) = 4 sin(2(x − C)) + 6 so we need to find C to
make (iii) true. Solve 8 = f (0) = 4 sin(2(0 − C)) + 6 or 1/2 = sin(−2C).
π
−2C = π/6) works so use C = −π/12. Answer: f (x) = 4 sin(2(x + 12
)) + 6
3.19 Each of the following is the graph of a trigonometric function. For each
one, give a possible formula for the function. Suggestion: Look for periods,
amplitudes and shifts.
3
2
1.5
1
0.5
2
1
-2
-1
1
-1
-2
-3
3
1
-2
Π
Π
5!2
Π Π3 Π!Π! ####
3 Π2 Π5 Π3 Π
!3!Π########
#### Π ########
########
! ######## -0.5
2
2
2
2
2
2
-1
-1.5
-2
10
7.5
5
2.5
2
Π
3Π
! ######## !Π ! ####
2-1
2
2
Π
####
2
3Π
Π ########
2
Π
3Π
####
! ######## !Π !-2.5
2
2
-5
-7.5
-10
Π
####
2
3Π
Π ########
2
Clockwise from upper left: y = 2 sin(2πx), y = − cos(x) − 1, y = tan x,
y = 2 sin(4x).
MTH 111 Exam III Practice Problems Answers/Hints
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3.20 Suppose that angles A, B both lie in quadrant II. Suppose cos A = −2/3
and sin B = 1/2. Find
a) sin(A) b) cos(2A)
c) sin(A + B).
Give exact answers in terms of radicals and fractions – no decimals.
q
q
√
a) sin A = + 1 − (−2/3)2 = 5/9 = 53. b) cos(2A) = cos2 (A) −
sin2 (A) = 4/9 − 5/9 = −1/9. c) Since B is in the second quadrant and
√
√
√
sin(B) = 1/2, cos(B) = − 3/2, so sin(A+B) = ( 5/3)(− 3/2)+(−2/3)(1/2).
3.21 Decide whether or not each of the following is a correct iidentity Answer
either YES or NO for each. Justify your answer algebraically. No credit without
such justification.
a) (cos α − 1)(cos α + 1) = sin2 α
b) 1 + sec2 (x) = tan2 (x)
c) sin(x − π/2) = − sin(x) cot(x)
d) cos4 (x) − sin4 (x) = cos(2x)
e)
sin(x + 4π)
= tan(x + π)
cos(x − 6π)
a) NO. (cos α − 1)(cos α + 1) = cos2 α − 1 = − sin2 α
b) NO. 1 + tan2 x = sec2 x.
c) YES. sin(x − π/2) = − sin(π/2 − x) = − cos(x) = − sin(x) cot(x)
d) YES.
cos4 (x) − sin4 (x) = (cos2 (x) − sin2 (x))(cos2 (x) + sin2 (x) = cos(2x) · 1
MTH 111 Exam III Practice Problems Answers/Hints
e) YES. sin(x + 4π) = sin x, cos(x − 6π) = cos x, tan(x + π) = tan x.
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