Math 150 Practice Solutions
1) A) cos 2 
1
 2  60   360  k , 2  300   360  k .
2
Divide both sides by 2 to get
  30 ,210 ,150 ,330



  30   180  k ,   150   180  k .
Let k = 0 and k =1,

B) 2 sin( 2 x )   3
Divide both sides by 2 to get sin( 2 x)  
3
.
2
Make triangles, one in QIII, the other in QIV, to get
4
5
2
5
 2k , 2 x 
 2k . Divide by 2, we get x 
 k , x 
 k . Pick k = 0 and k = 1,
3
3
3
6
2 5 5 11
x
,
,
,
3 3 6 6
2x 
c)
cos(3x)  
1
2
Make two triangles, one in QII, the other in QIII, 3 x 
get x 

2
5 2
 k , x 
 k .
4 3
12 3
3
5
 2k , 3x 
 2k . Divide by 3, we
4
4
d) cos( 2 x)  1, 0  x  2 :
1
First find the angle whose cosine is -1: this angle is
2x    2k . x 

2
 (x is -1 on the negative x-axis).
 k . Pick k  0 and k  1 , we get
 3
2
,
Solve
2
2) First sketch the graph of r  1 2 sin  in the Cartesian coordinates. It is important that you
1
find the x-intercepts by setting r = 0. 0  2 sin   1  sin       210,330 . Since
2

r  1  2 sin   1 when   0 , r  1  2 sin   3 when   90 and r  1  2 sin   1when   180  ,


the graph goes through (1,0), (3,90 ), (1,180 ) in polar coordinates and returns to the origin at
210 degrees (since r = 0 at 210). There is a loop from 210 to 330 on the opposite sides of
210 to 330. Since 210 to 330 is from QIII to QIV, the loop on the opposite side would be
from QI to QII. Then after exiting the loop at 330, since r
graph goes back to
3) A) cos  
(1,360  ) .
3(1)  (1)( 2)
(3)  (1)
2
 1  2 sin   1 when   360  , the
2
(1)  2
2
2
   cos 1 (
1
50
)  98.2  b) if two vectors are
perpendicular, then their dot product must be 0. (1)( 1)  5( x)  0  x 
1
5
c) Since it is not a nice triangle, you must use a calculator:
x  4 cos 200   3.8, y  4 sin 200   1.4 Thus (4.200  )  (3.8,1.4)
d)
r  32  2 2  13 .
For the angle, first find the reference angle:
Since the angle is in QII,
  180  56.3  236.3 .
Thus it is
5ˆ  tan 1
3
 56.3 .
2
( 13, ,236.3 )
2
x
100

 x  291 . y  291sin 46   209


sin 116
sin 18
4) See the picture: First find x:
5) Construct two triangles: one in QIII, the other in QIV
1
7
11
7 2
sin 3    3 
 2k , 3 
 2k . Divide both sides by 3 , we get  
 k ,
2
6
6
18 3
11 2
7 19 31 11 23 35

 k and use k = 0, 1, 2.  
,
,
,
18 3
18 18 18 18 18 18
3
5
7
 2 
 2k , 2 
 2k . Then dividing by 2, we get
6) A) cos( 2 )  
2
6
6
5
7

 k ,  
 k
12
12
b) tan( 3 )  1  3 
7) x  2 cos t  cos t 
3
7
 2
7 2
 2k , 3 
 2k  3   k ,  
 k
4
4
4 3
12 3
x
y
, y  3 sin t  sin t 
2
3
2
. Using
2
 x  y
cos 2 t  sin 2 t  1, we get       1 , an
2  3
ellipse. Make a table to find its orientation,
T
x  2 cos t
0
2
0

2
y  3 sin t
0
3
3

-2
a)
0
x  2  cos t , y  1  sin t  cos t  x  2, sin t  y  1  ( x  2) 2  ( y  1) 2  1 , a circle with center
(2,1), radius1.
T
x  2  cos t
y  1  sin t
0
3
2
1
0
1
1

2

8) A) First multiply both sides by r to get r 2  6r cos   x 2  y 2  6 x  x 2  6 x  y 2  0 To
6 2
)  9 to both sides: x 2  6 x  9  y 2  9  ( x  3) 2  y 2  32 .
complete the square, add (
2
2
2
2
This is a circle of radius 3 centered at (3.0). B) x  y  4 y  r  4r sin   r  3sin  by
dividing both sides by r.
9)
Recall that plane vector+ wind vector= actual vector, the angle=450- the heading. Plane =
 210 cos( 293), 210 sin( 293)  82.1,193.3  Wind =  61cos(310), 61sin( 310)  39.2,46.7 
Plane + wind =  82.1,193.3    39.2,46.7  121.3,240  . The actual speed is
240
121.3 2  (240) 2  269 mph. For the true course, the reference angle is tan 1 (
)  63.2 .
121.3
Notice that the vector is in the 4th quad. Therefore, the bearing is 90+63.2 = 153.2
10)
a) Since x=-2, y=2, construct a 1  1  2 triangle in the 2nd quadrant. Multiply by
the radius ( 2 2, ,
3
)
4
2 to het
11

in QIV, and the reference angle is
, construct a short 2  1  3 triangle. Y
6
6
half of the radius. For x, multiply by 3
b) Since
(4,
11
)  (2 3,  2)
6
11) A) First sketch the graph of r  cos 2 from 0 to 360 in the Cartesian coordinates. The xintercepts are 45 degrees ,135 degrees, 225 degrees, and 315 degrees. There is a half loop
from 0 to 45, complete loops from 135 degrees to 225 degrees, and from 225 degrees to
315 degrees, then a half loop from 315 to 360.. From 0 to 45: QI (same side) since r is
4
positive. From 45 to 135: QIII and QIV (opposite side) since r is negative: From 135 to 225
QII and QIII (same side) since r is positive: From 225 to 315 QI and QII (opposite) since r is
negative: From 315 to 360 QIV(same side) since r is negative
B) First sketch the graph of r  sin 2 from 0 to 360 in the Cartesian coordinates. The xintercepts are 0 degree ,90 degrees, 180 degrees, and 270 degrees. There are loops from
0degree to 90 degrees , 90 degrees to 180 degrees, 180 degrees to 270 degrees, and from
270 degrees to 360 degrees. From 0 to 90: QI (same side) since r is positive. From 90 to
180: QIV (opposite side) since r is negative: From 180 to 270 QIII (same side) since r is
positive: From 270 to 360 QII (opposite) since r is negative:
5
b) First sketch the graph of r  sin 3 from 0 to 360 in the Cartesian coordinates. The xintercepts are 0 degree 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300
degrees and 360 degrees. The loop from 0 to 60 is in QI (same side) since r is positive.
From 40 to 120: sketch the graph in QIII, then QI V(opposite side) since r is negative:
From 120 to 180 sketch the graph in QIII (same side) since r is positive: From 180 to
240 QI (opposite) since r is negative. But you will notice at this point that it starts
repeating. So you may stop here, obtaining a graph of 3 loops
\\
6
d) r  4 sin 2  r  2 sin 2 First sketch the graph of r  4 sin 2 from 0 to 360 in the
Cartesian coordinates (2 cycles) . The x-intercepts are 0 degrees, 90 degrees, 180 degrees,
270 degrees, and 360 degrees. However, we only take the positive portion of the graph
since r is squared. Thus we throw away the negative portion from 90 to 180, 270 to 360.
Then we take plus/minus of square root. Doing so would alter the shape of the sine curve
slightly, but not enough for our rough sketch of the graph. Its maximum height now
becomes 2. From 0 to 90, there are two curves: one positive, the other negative. Another
two will be from 180 to 270, thus forming two “eggs”. In polar coordinates, the two sine
curves from 0 to 90 will make two loops: one in QI (positive) and the other in QIII
(negative). Then from 180 to 270, it repeats.
2
2
7
e) r  9 cos 2  r  3 cos 2 First sketch the graph of r  9 cos 2 from 0 to 360 in the
Cartesian coordinates. The x-sintercepts are 45 degrees ,135 degrees, 225 degrees, 315
degrees, 360 degrees. However, we only take the positive portion of the graph since r is
squared. Throw away the portion from 45 to 135, 225 to 315. Then we take plus/minus of
square root. Doing so would alter the shape of the cosine curve slightly, but not enough to
warrant any consideration. From 0 to 45, there are two curves: one positive, the other
negative. Another two will be from 135 to 225 and two more from 315 to 360. In polar
coordinates, the two cosine curves from 0 to 45 will make two half-loops: one in QI
(positive) and the other in QIII (negative). Then from 135 to 225, thus loop is completed.
2
2
8
12) A) First sketch the graph of r  1 2 cos in the Cartesian coordinates. It is important that
1
you find the x-intercepts by setting r = 0. 0  2 cos   1  cos       120,240 . Since
2



90
when
,
when
,
the
graph
goes through
r  1  2 cos  3
  0 r  1  2 cos  1

(3,0), (1,90 ) in polar coordinates and returns to the origin at 120 degrees (since r = 0 at 120).
There is a loop from 120 to 240 on the opposite sides of 120 to 240. Since 120 to 240 is
from QII to QIII, the loop on the opposite side would be from QII to QI. Then after exiting
the loop at 240, since
graph passes through
r  1  2 cos  1 when   270  and r  1  2 cos  3 when   360  , the
(1,270  ), (3,360  ) .
9
B) First sketch the graph of
r  1 2 cos
in the Cartesian coordinates. It is important that
you find the x-intercepts by setting r = 0. 0  1  2 cos  cos  
1
   60,300 . By looking at
2
its Cartesian graph, r is negative between 0 and 60 and between 300 and 360. Since
r  1  2 cos  1 when   0 , and back to 0 at 60 degrees, this portion of the graph is in the
third quadrant. r  1  2 cos  1 at 90 degrees, r  1  2 cos  3 at 180 degrees,
r  1  2 cos  1 and when   270 
the graph passes through
(1,90  ), (3,180 ), (1, 270  ) . Then it
goes back to the origin at 300 degrees. Then the graph is in the second quadrant from 300
and 360 since r is negative, completing the inner loop.
10
13) x  2  3 cos t , y  1  3 sin t  x  2  3 cos t , y  1  3 sin t
(3 cos t ) 2  (3 sin t ) 2  9  x  2   y  1  3 2 . This is a circle of radius 3, centered at (2,-1)
2
2
T
x  2  3 cos t
y  1  3 sin t
0
5
2
-1
2
-1
-1

2

14)
x
25
y
25

,

sin 25 sin 110 sin 55 sin 110
x  11 mi, y  21mi
15) Factor the expression as (cos 3  3)( 2 cos 3  1)  0  cos 3  
1
,3 . But cos 3  3 is
2
impossible.
3  120   360  k    40   120  k , 3  240   360  k    80   120  k
11
16) A)
x  4 cos110  , y  4 sin 110   (1.4,3.8) B) divide by its magnitude
  cos 1 (
 1,3  4,5 
1  9 16  25
)  cos 1 (
19
410


3 3
17) A) x  3 cos , y  3 sin
(
,
)
4
4
2 2
B) r 
10 : 
1
10
,
3
10
 c)
)  20.2 
x 2  y 2  3  1  2 . Make a triangle with x   3 , y   1 . It is a short 30-60-90
triangle in QIII with the hypotenuse 2, the angle 210.
(2,210  )
1
2
3
) :Or use a triangle. d) Make a triangle with
2
x  1, y   1 . This is a 45-45-90 triangle in QIV with r = 2 with the angle 315. ( 2 ,315  )
c) x  ( 2) cos 120, y  ( 2) sin 120  ( ,
d) First multiply both sides by r to get
complete the square, add (
r 2  2r sin   x 2  y 2  2 y  x 2  y 2  2 y  0 To
2 2
)  1 to both sides: x 2  y 2  2 y  1  1  x 2  ( y  1) 2  12 .
2
This is a circle of radius 1 centered at (0,1).
18) Since it is not a nice triangle, you must use a calculator:
x  4 cos 20   3.8, y  1.4 Thus
(4.20  )  (3.8,1.4)
19) r
 12  ( 2 ) 2  5 .
For the angle, first find the reference angle:
  180  54.7  125.3 .
(ˆ  tan 1 2  54.7 .
Since
( 5 ,125.3 )
v
5 12
2
2
  , 
20) Divide by its magnitude: Since  5,12   5  12  13,
v
13 13
4
1  2,4  4,1 
21)   cos (
)  cos 1 (
)  77.5
4  16 16  1
340
2
22) a cos 2  3 cos 2  2  0 . This can be factored as
cos 2 2  3 cos 2  2  (cos 2  1)(cos 2  2)  0  cos 2  1, cos 2  2 But cos 2  2 is
impossible. Solve cos 2  1 . We know (or I know) that the cosine is 1 on the positive xthe angle is in QII,
Thus it is
axis, the angle is 0.
2  0  2k    k
Pick k = 0 and k = 1,
  0, 
12
c) tan 2  1 : We first take plus/minus of square root to solve for tan 2 : tan 2  1 . First
to solve tan 2  1, construct one triangle each in QI and QIII with the reference angle of
45.
2

5
 2k Divide both sides by 2 (or multiply by ½) ,
4
4

5
 9 5 13
   k ,  
 k Pick k = 0 and k =1,   ,
,
. Next solve tan 2  1 by
8
8
8 8 8 8
tan 2  1  2 
 2k ,2 
constructing two triangles, one in QII and the other in QIV (since the tangent is
negative)with the reference angle of 45. tan 2  1  2 
3
7
 2k , 2 
 2k . Divide
4
4
3
7
 k ,  
 k . Pick k = 0 and k =1, ,
8
8
3 11 7 15
 9 5 13 3 11 7 15

,
,
,
,
,
,
,
,
,
Therefore,   ,
8 8
8 8
8 8 8 8 8 8 8
8
both sides by 2 (or multiply by ½) ,

D) First divide by 2, then take plus/minus of square root. We get
solve
45.
sin 2  1/ 2 . First to
sin 2  1 / 2 , construct one triangle each in QI and QII with the reference angle of

3
 2k Divide both sides by 2 (or multiply by ½) ,
4
4

3
 9 3 11
   k ,  
 k Pick k = 0 and k =1,   ,
,
. Next solve sin 2  1 / 2 by
8
8
8 8 8 8
sin 2  1 / 2  2 
 2k ,2 
constructing two triangles, one in QIII and the other in QIV (since the sine is negative)with
5
7
 2k , 2 
 2k . Divide both sides
4
4
5
7
5 13
 k ,  
 k . Pick k = 0 and k =1, ,  
,
by 2 (or multiply by ½) ,  
,
8
8
8 8
7 15
 9 5 13 3 11 7 15
,
,
,
,
,
,
,
Therefore,   ,
8 8
8 8 8 8 8 8 8
8
the reference angle of 45. sin 2  1 /
2  2 
13
23)
remember, plane + wind = actual, the angle=450- the bearing.
Plane =
 500 cos(310), 500 sin( 310)  321,383  Wind =  80 cos( 408), 80 sin( 408)  54, 59  Plane +
wind =  321,383    54, 59  375,324  . The actual speed is
1
For the true course, the reference angle is tan (
375 2  (324) 2  496 mph.
324
)  40.8 . Notice that
375
the vector is in the 4th quad. Therefore, the true course is
=450-angle =450-319.2=130.8 degrees
  360  ˆ  319.2 .
The bearing
sin 

 1  tan   1   
cos 
4
2
2
2
2
c) r  4 cos  : Multiply both sides by r to get r  4r cos  Since we know x  y  r , x  r cos 
24) a) r  3 
x 2  y 2  3  x 2  y 2  9 b) x  y  r cos   r sin  
, r  4r cos  becomes
2
x 2  y 2  4x
25) wind= actual-plane, the angle=450- the heading. Actual =
 464 cos(199), 464 sin( 199)  438.7,151.1  Plane =
 460 cos( 202), 460 sin( 202)  426.5,  172.3  Actual-plane =
 438.7,151.1    426.5,  172.3  12.2, 21.1  . The wind speed is
(12.2) 2  (21.1) 2  24.4 mph. For the direction of the wind, the reference angle is
21.1
tan 1 (
)  60 . Notice that the vector is in the second quad. Therefore, the heading of the
12.2
0
wind is 270  60  330 .
14
26) A) To find the speed, observe that the angle between the plane vector and actual vector is
10 degrees. . x 
500 2  490 2  2(500)( 490) cos 10  86.9 mph. For the direction, first find the
sin  sin 10
angle opposite 490.

   78.2 or   101.8 . However, the angle opposite 500 mph
490
86.9
is the largest in the triangle. If the angle opposite 490 is 101.8, then the angle opposite 101.8
(the largest one) is greater than 101.8. This is clearly impossible. Therefore, the angle
opposite 490 mph is 78.2. See the picture below. Use the alternate interior angles, the angle
A is 50 degrees. The angle B is 78.2-50=28.2. Therefore, the heading of the wind is
180+28.2=208.2 degrees. B) Vector approach: remember, wind= actual-plane, the angle=450the heading. Actual =  490 cos(150), 490 sin( 150)  424.4, 245  Plane =
 500 cos(140), 500 sin( 140)  383, 321.4  Actual-plane =
 424.4, 245    383, 321.4  41,  76.4  . The wind speed is (41) 2  (76.4) 2  86.7 mph.
1 76.4
)  61.8 . Notice that the vector
For the direction of the wind, the reference angle is tan (
41
15