MATH 32 SPRING 2013
FINAL EXAM SOLUTIONS
(1) (9 points) Fill in the following table:
Solution:
π/6
5π/4
sin
cos
√
1/2
√
√3/2
−
−
√ 2/2
√ 2/2
√
√
2− 2
2
π/8
2+ 2
2
tan
√
√
1/ 3 = 3/3
√ √ 1 √
√2−√2 = 2 − 1
2+ 2
π/6 is one of our special angles - we know its sine and cosine. If you don’t have its
tangent memorized, it can be found by dividing sine by cosine.
5π/4 is related to the angle π/4, but in the third quadrant, so it has negative sine and
cosine. Again, its tangent can be found by dividing sine by cosine.
For π/8, we need to use the half-angle formulas. Note that since π/8 is in the first
quadrant, it has positive q
sine and cosine.
q
√
q
√
1− 2/2
=
q 2√
q
1+cos(π/4)
cos(π/8) = cos( π/4
= 1+ 2 2/2 =
2 ) =√
2
√ √
√
sin(π/8)
2
2
√
√2−√2 .
= 2−
·
=
tan(π/8) = cos(π/8)
√
2
2+ 2
2+ 2
sin(π/8) = sin( π/4
2 )=
1−cos(π/4)
2
=
√
2− 2
=
q 4√
2+ 2
=
4
√
2− 2
√2 √
2+ 2
2
The above is a fine answer for tangent, but you can get a nicer answer if you rationalize
the√denominator:
√ √ √ √
√ √
√
√
√
√
√
√ 2 = √2 − √2 =
√2−√2 = (√2−√2)(√2−√2) = √(2−√2)(2−√2) = 2−
2 − 1.
4−2
2
2
2+ 2
(
2+ 2)(
2− 2)
(2+ 2)(2− 2)
(2) Consider the rational function
f (x) =
x2 − 1
x2 + 1
(a) (3 points) Find the x-intercepts of f or explain why it has none.
(b) (3 points) Find the horizontal asymptote of f or explain why it has none.
(c) (3 points) Find the vertical asympotes of f or explain why it has none.
Solution:
(a) (−1, 0) and (1, 0). The x-intercepts of f are at the zeros of the numerator. x2 − 1 =
(x − 1)(x + 1), which has zeros at x = ±1.
(b) y = 1. The leading terms of the numerator and denominator are x2 and x2 (same
2
degree). Their ratio is xx2 = 1, so f has a horizontal asymptote at y = 1.
(c) None. The vertical asymptotes of f are at the zeros of the denominator. But x2 + 1
has no zeros, so f has no vertical asymptotes.
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MATH 32 SPRING 2013 FINAL EXAM SOLUTIONS
(3) Consider the right triangle below.
(a) (3 points) Find cos(A).
(b) (3 points) Find sin(A).
(c) (3 points) Find cos(A + π/4).
Solution:
(a) √
2/3. In a right triangle, the cosine of an angle is adjacent/hypoteneuse = 2/3.
(b) 5/3. In a right triangle, the sine of an angle is opposite/hypoteneuse√= b/3. To find
b, use the Pythagorean Theorem: 22 + b2 = 32 , b2 = 9 − 4 = 5, so b = 5.
(c) The angle sum formula
for cosine√tells √us that√ cos(A
+ π/4) = cos(A) cos(π/4) −
√
√ √
√
5 2
2 2
10
2 2− 10
2 2
.
sin(A) sin(π/4) = 3 2 − 3 2 = 6 − 6 =
6
(4) (9 points) Does the absolute value function f (x) = |x| have an inverse? Explain why or
why not.
Solution: No, it does not have an inverse - it does not pass the horizontal line test.
(5) (9 points) Find all values of x satisfying 2x+1 = 3x .
Solution: There are several ways to write the answer, depending on which logarithm
you took. Here are two possibilities:
Take log2 of both sides:
log2 (2x+1 ) = log2 (3x )
x + 1 = x log2 (3)
x log2 (3) − x = 1
x(log2 (3) − 1) = 1
x=
1
log2 (3) − 1
Or, take ln of both sides:
ln(2x+1 ) = ln(3x )
(x + 1) ln(2) = x ln(3)
x ln(2) + ln(2) = x ln(3)
x ln(3) − x ln(2) = ln(2)
x(ln(3) − ln(2)) = ln(2)
x=
ln(2)
ln(3) − ln(2)
MATH 32 SPRING 2013
FINAL EXAM SOLUTIONS
3
(6) Here are three quick questions about the inverse cosine function.
(a) (3 points) What are the domain and range of cos−1 ?
(b) (3 points) Evaluate cos(cos−1 ( 43 )).
(c) (3 points) Evaluate cos−1 (cos(− π7 )).
Solution:
(a) Domain: [−1, 1], Range: [0, π].
(b) 3/4. cos−1 (3/4) is an angle with cosine 3/4, so cos(cos−1 (3/4)) = 3/4.
(c) π/7. Note that −π/7 is not in the range of cos−1 , so cos−1 (cos(−π/7)) 6= −π/7.
Instead, it is the angle in the range [0, π] with the same cos as −π/7, namely π/7.
(7) Let L1 be the line which passes through the points (1, −2) and (3, 4). Let L2 be the line
perpendicular to L1 which passes through the origin.
(a) (3 points) Find an equation for L1 .
(b) (3 points) Find an equation for L2 .
(c) (3 points) Find the point of intersection of L1 and L2 .
Solution:
6
(a) Slope: 4−(−2)
3−1 = 2 = 3. Equation: In point-slope form, y − 4 = 3(x − 3). Solving for y
and simplifying, y = 3(x − 3) + 4 = 3x − 9 + 4 = 3x − 5.
(b) Slope: − 13 , since it is perpendicular to L1 . Equation: y = − 13 x, since it goes through
the origin (0, 0).
(c) To find the point of intersection, solve the equations y = 3x − 5 and y = − 13 x simultaneously. Setting the two expressions for y equal to each other,
1
3x − 5 = − x
3
1
3x + x = 5
3
10
x=5
3
3
15
3
x=5·
=
= .
10
10
2
1
1 3
1
To find y, plug in our value for x: y =
−
x
=
−
3
3 · 2 = −2.
So the point of intersection is − 32 , 21 .
√
(8) In the triangle below (not drawn to scale), let A = 45◦ , B = 30◦ , and a = 2.
(a) (3 points) What is the measure of angle C?
(b) (6 points) What is the length of side b?
Solution:
(a) We know that A + B + C = 180◦ , so C = 180◦ − 45◦ − 30◦ = 105◦ .
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MATH 32 SPRING 2013 FINAL EXAM SOLUTIONS
(b) Use the Law of Sines:
sin(A)
sin(B)
=
a√
b
2
1
√2 = 2
b
2
Cross-multiplying,
√
1√
2
2√
2 2
b=
·√
2
2
b=1
2
b=
2
(9) (9 points) Show that for all angles θ,
sin(θ) cos(θ) tan(θ) + sin(θ) cos(θ) cot(θ) = 1
Solution:
sin(θ) cos(θ) tan(θ) + sin(θ) cos(θ) cot(θ) = sin(θ) cos(θ)
cos(θ)
sin(θ)
+ sin(θ) cos(θ)
cos(θ)
sin(θ)
= sin2 (θ) + cos2 (θ)
=1
(10) (9 points) Evaluate log2 (8101 ).
Solution: log2 (8101 ) = 101 log2 (8) = 101 · 3 = 303.