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Sample Midterm Exam - SOLUTIONS
1. Find the exact value for each of the following expressions.
p
2
sin 45 sin 30 =
2
(a) cos 75 = cos (45 + 30 ) = cos 45 cos 30
p
p
3
2
2 1
=
2 2
p
6
p
2
4
(b) tan 22:5 =
Solution: First we will …nd cos 22:5 : We will use the double angle formula for cos 2x, where
x = 22:5 .
cos p
45
2
p 2
2
+1
p2
2
+1
2
vp 2
u
u 2
t
+1
2
2
= 2 cos2 22:5
1
= 2 cos2 22:5
1
= 2 cos2 22:5
= cos2 22:5
= cos 22:5
Since the angle 22:5 is in the …rst quadrant, our result is the positive solution. We simplify
the answer
vp
vp
vp
u
u
u
sp
u 2
u 2 2
u 2+2
pp
t
t
+1 t
+
2
+
2
2+2
2
2
2 =
2
cos 22:5 =
=
=
=
2
2
2
4
2
We can now …nd sin 22:5 :
sin2 22:5 + cos2 22:5
sin2 22:5
sin 22:5
sin 22:5
= 1
= 1 cos2 22:5
p
=
1 cos2 22:5
=
p
1
v
u
u
cos2 22:5 = t1
pp
2+2
2
We simplify the answer
v
s
s
!2 s
pp
u
p
p
u
4
2+2
2+2
4
2+2
sin 22:5 = t1
= 1
=
=
2
4
4
4
s
s
s
p
p
p
p
p
4
2+2
4
2 2
2
2
2
2
=
=
=
=
4
4
4
2
Now
p
p
!2
p
2+2
=
4
s
2
p
p
p
p
p
2
2
2
2
sin 22:5
2
2
2
2
p
pp
=
= pp
tan 22:5 =
= pp
=
cos 22:5
2
2+ 2
2+2
2+2
2+2
2
2
1
This is acceptable as a …nal answer. However, if we rationalize the denominator under the
square root, we …nd a simpler expression.
s
s
s
s
p 2
p 2
p
p
p
2
2
2
2
2
2
2
2 2
2
p =
p
p =
tan 22:5 =
=
=
4 2
2
2+ 2
2+ 2 2
2
q
p 2
p
2
2
2
2
p
= p
=
2
2
Then, when we rationalize this new denominator, we get that
p
p
p p
2 2
2
2
2 2 2
2
p =
tan 22:5 = p
=
2
2
2
2
1
=
p
2
1.
Note: there are shorter ways to solve this problem. There are various half-angle formulas for
x
tan presented in the text book.
2
p
3
(c) cos 68 sin 8
sin 68 cos 8 = sin (8
68 ) = sin ( 60 ) = sin 60 =
2
2
tan
+ tan
p
2
2
3
5
15
5
(d)
= tan
+
= tan
+
= tan
= tan = 3
2
15
5
15
15
15
3
1
tan
tan
15
5
2. Prove each of the following identities.
cot2 x 1
2 cot x
(a) cot 2x =
Solution:
cos2 x sin2 x
cos2 x sin2 x
cos2 x
1
2
cot2 x 1
sin2 x
sin2 x sin2 x =
x
=
= sin cos
=
RHS =
cos x
x
2 cos x
2 cot x
2
2
sin x
sin x
sin x
2
2
2
2
cos x sin x sin x
cos x sin x
cos 2x
=
=
=
= cot 2x = LHS
2
2 cos x
2 sin x cos x
sin 2x
sin x
(b) 4 sin4 x = 1
Solution:
2 cos 2x + cos2 2x
RHS = 1
= 1
(c) cos 3x = 4 cos3 x
Solution:
LHS =
=
=
=
=
2 cos 2x + cos2 2x = 1 2 1 2 sin2 x + 1 2 sin2 x
2 + 4 sin2 x + 1 4 sin2 x + 4 sin4 x = 4 sin4 x = LHS
2
=
3 cos x
cos 3x = cos (x + 2x) = cos x cos 2x sin x sin 2x =
cos x 2 cos2 x 1
sin x (2 sin x cos x) =
3
2 cos x cos x 2 sin2 x cos x = 2 cos3 x cos x 2 1
2 cos3 x cos x 2 cos x 1 cos2 x = 2 cos3 x cos x
4 cos3 x 3 cos x = RHS
2
cos2 x cos x =
2 cos x + 2 cos3 x =
(d) tan 3x = tan x
2 cos 2x + 1
2 cos 2x 1
Solution:
tan x + tan 2x
LHS = tan 3x = tan (x + 2x) =
=
1 tan x tan 2x
2 tan x
1 tan2 x
2 tan x
tan x
1 tan2 x
tan x +
1
=
2
2
1+
2
1 tan x
1 tan2 x
= tan x
=
2 tan x
2 tan x
1 tan x
1 tan x
1 tan2 x
1 tan2 x
1
0
2
1+
B
1 tan2 x C
1 tan2 x + 2
1 tan2 x
C = tan x
=
tan x B
@
2 tan x
1 tan2 x A
1 tan2 x tan x (2 tan x)
1 tan x
1 tan2 x
3 tan2 x
3 tan2 x
tan x
=
tan
x
=
1 tan2 x 2 tan2 x
1 3 tan2 x
sin2 x
3
2x
3 cos2 x sin2 x
cos2 x
cos
= tan x
tan x
=
sin2 x cos2 x
cos2 x 3 sin2 x
1 3 2
cos x
2 cos2 x 2 sin2 x + cos2 x + sin2 x
tan x
=
2 cos2 x 2 sin2 x cos2 x sin2 x
2 cos2 x sin2 x + cos2 x + sin2 x
2 cos 2x + 1
tan x
=
tan
x
= RHS
2
2
2 cos 2x 1
2 cos2 x sin x
cos2 x + sin x
tan x 1 +
=
=
=
=
=
=
3. Find the exact value of all solutions for each of the following equations. Present your answer in
radians.
(a) 2 + 3 sin x = cos 2x
Solution: cos 2x = 1
2 sin2 x. And so
2 + 3 sin x
2
2 sin x + 3 sin x + 1
(sin x + 1) (2 sin x + 1)
sin x + 1
=
=
=
=
1 2 sin2 x
0
0
0 or 2 sin x + 1 = 0
giving us two sets of solutions:
sin x + 1 = 0
sin x =
1
x =
+ 2k where k is an integer
2
3
or
2 sin x + 1 = 0
1
2
sin x =
+ 2k where k is an integer and
6
5
+ 2k where k is an integer.
6
x =
x =
So the set of all solutions is
x =
x =
x =
2
+ 2k
+ 2k
6
5
+ 2k where k is an integer.
6
(b) sin 2x = 2 cos x
Solution:
sin 2x
2 sin x cos x
2 sin x cos x 2 cos x
2 cos x (sin x 1)
cos x
=
=
=
=
=
2 cos x
2 cos x
0
0
0 or sin x
1=0
giving us two sets of solutions:
cos x = 0
x =
+ k where k is an integer.
2
or
sin x 1 = 0
sin x = 1
x =
+ 2k where k is an integer.
2
The second solution set is already contained in the …rst one. So all solutions together are
x = + k where k is an integer.
2
p
3
(c) cos 3x =
2
Solution:
p
3
cos 3x =
2
3x =
150 + k 360 where k is an integer
x =
50 + k 120 where k is an integer.
4
Finally, we convert the answer to radians
50
x =
5
2
+k
where k is an integer.
18
3
180
+ k 120
where k is an integer
x =
180
8
4. Suppose that sin =
and is not in the fourth quadrant; cos
17
quadrant. Find the exact value for each of the following.
=
Solution: The conditions given place into the third quadrant and
These will determine the signs of the other trigonometric functions.
Since
is not in the …rst
into the fourth quadrant.
is in the third quadrant, cos
cos
Similarly, since
=
is negative. For the rest,
s
p
8
1 sin2 =
1
17
12
and
13
is in the fourth quadrant, sin
sin
=
q
1
2
cos
2
=
15
17
is negative. For the rest,
s
2
12
5
=
1
=
13
13
Now we have all we need:
sin
=
cos
=
tan
=
8
17
15
17
8
15
5
13
12
=
13
5
=
12
sin
=
cos
tan
Using the calculator, we can also come up with approximate values for
a practical way to check our solution.
8
= arcsin
= 208: 072 486 935 8
17
= arccos
12
13
=
and . These will give us
22: 619 864 948 04
(a) tan (
)=
Solution:
tan (
tan
tan
)=
1 + tan tan
=
8
15
1+
8
15
5
5
12
5
12
8 (4) + 5 (5)
57
19
60
=
= 60 =
2
7
20
1
9
9
9
7
=
171
140
171
= 1: 221 43 and
140
( 22: 619 864 948 04 )) = 1: 221 43 and compare the decimals.
We can use the calculator to check our result by entering the fraction
tan (208: 072 486 935 8
(b) cos ( + ) =
Solution: cos ( + ) = cos cos
sin sin
15
17
=
12
13
8
17
5
13
=
220
221
(c) sin 2 =
Solution: sin 2 = 2 sin
(d) sin
2
=2
8
17
15
17
=
240
=
289
0:830 449 826 989 6
=
v
u
u 1 + 15
t
17 = p4 = 0:970 142 500 145 3
Solution:
2
17
5. Find the exact value of tan
if
is the acute angle formed by the lines 2x 3y = 5 and 5x + 3y = 1.
Solution: Find the slopes of the lines …rst. We do this by solving the equations for y and then read
the coe¢ cients of x.
2x 3y = 5
2x 5 = 3y
2
y =
x
3
5
3
and
5x + 3y = 1
3y =
5x + 1
5
1
y =
x+
3
3
So the slopes are m1 = tan
tan (
=
2
and m2 = tan
3
=
tan
tan
)=
1 + tan tan
5
. We are now looking for tan (
3
5
3
=
1+
2
3
5
3
2
3
).
= 21
note: if we accidentally compute tan (
), which is the tangent of the obtuse angle, the result
would be 21, since the tangents of supplemental angles are opposites of each other. Then we
would just have to conclude from that that the answer is 21.
6. Find the area of a regular polygon of 10 sides, inscribed in a circle of radius 5 m:
Solution: Let us …nd the area of one triangle …rst. Then we will multiply that area by 10 sice 10
such triangles make up the polygon.
6
Let us label the base, AB by x, and the height, OM by y. Then OBM triangle is a right triangle,
with M OB\ = 18 , and the hypotenuse is 5 m.
x
x
sin 18 = 2 =
5
10
x = 10 sin 18
To …nd y:
y
5
y = 5 cos 18
cos 18
=
Thus the triangle AOB has area
1
1
xy = (10 sin 18 ) (5 cos 18 )
2
2
= 25 sin 18 cos 18
A =
Thus the polygon’s area is
10A = 10 (25 sin 18 cos 18 ) = 250 sin 18 cos 18 =
= 73: 473 16 m2
7