MATH 147
Quiz Key #1
1/28/2016
(1) Find the equation of the line passing through (5, −1) and perpendicular to
the line passing through (−2, 1) and (1, −2) . Express your answer in slopeintercept form.
Solution: First, we find the slope of the line passing through (−2, 1) and
(1, −2) . We will call it m perp since it is the slope of the line perpendicular to
the line in which we are interested.
m perp =
−2 − 1
−3
y2 − y1
=
=
= −1.
x2 − x1
1 − (−2)
3
The slope of the line in which we are interested is the negative reciprocal of
m perp :
1
m=−
= 1.
m perp
Now, we plug into point-slope form as follows:
y − y1 = m ( x − x1 ) .
y − (−1) = 1 ( x − 5) .
y + 1 = x − 5.
y = x − 6.
1
(2) Find the exact
values of the following expressions:
(a) sin − 5π
4
(b) cos 5π
6
Solution: (a) We start by drawing our angle in the unit circle and sketching the
relevant triangle [see the figure below on the left]. Then
5π
opp.
1
sin −
=
= √ .
4
hyp.
2
(b) We start by sketching the unit circle with relevant information [see the
figure above on the right]. Then
√
5π
3
cos
=−
.
6
2
2
(3) Solve 2 cos θ sin θ = sin θ on [0, 2π ) .
Solution:
2 cos θ sin θ = sin θ.
2 cos θ sin θ − sin θ = 0.
sin θ (2 cos θ − 1) = 0.
sin θ = 0 or 2 cos θ − 1 = 0.
1
sin θ = 0 or cos θ = .
2
We are first interested, therefore, in the angles on the unit circle where the ycoordinate is 0. This happens at the points on the x-axis which are at the angles
0, π, 2π. However, 2π is not in our domain. Therefore, the solutions to sin θ = 0
are given by θ = 0 or π. Next, we are interested in where the cos θ = 12 . One
such angle is π3 . Therefore, since cosine is positive in Quadrants I and IV, we
need to find every angle in Quadrants I and IV with a reference angle of π3 .
1
π
5π
Since these are π3 , 5π
3 , the solutions to cos θ = 2 are given by θ = 3 or 3 .
Therefore, the solutions to the original equation are given by
θ = 0, π,
π
5π
, or
.
3
3
(4) Completely simplify the expression
32 · 31/2
.
3−1/2
Solution: Since the two factors in the numerator and the expression in the
denominator have the same base, we can simplify as follows:
32 · 31/2
32+(1/2)
= −1/2 = 32+(1/2)−(−1/2) = 32+(1/2)+(1/2) = 33 = 27.
−
1/2
3
3
3