Sample Midterm Exam - SOLUTIONS

Sample Midterm Exam - SOLUTIONS

1. Find the exact value for each of the following expressions.

(a) cos 75 = cos (45 + 30 ) = cos 45 cos 30

pp 23

sin 45 sin 30 = 22

p

pp

21 6 2

=

22

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 p45 = 2 cos2 22:5 1

2 = 2 cos2 22:5 1

p2

2 +1

=

2 cos2 22:5

p2

2

+1

2

= cos2 22:5

v u u t

p2 2 +

1

2

= cos 22:5

2

Since the angle 22:5 is in the ...rst quadrant, our result is the positive solution. We simplify

the answer

cos 22:5

=

v u u t

p 2

2

+1

=

v u u t

p 2

2

+

2 2

=

v u u t

p 2

+

2

2

=

sp 2

+

2

=

pp 2

+2

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

= p1 cos2 22:5

=1 p

=1

cos2 22:5 cos2 22:5

= v u u t1

pp

!2

2+2

2

We simplify the answer

sin 22:5 = v u u t1

pp

!2 s

2+2

=1

2

p

s

2+2 4

=

4

4

p

s

2+2 4

=

4

s 4

=

p

sp

s p pp

2+2

4 22 2 2 2 2

=

=

=

4

4

4

2

p 2+2 = 4

Now sin 22:5

tan 22:5 = cos 22:5

pp

2 2pp

= pp2 = 2+2

2 2

2

2

pp sp

pp2

= p2p

2 =

2+2

2+2

2 p2 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.

tan 22:5

sp sp

2 2 2 22

=

p=

p

=

q2 +

2 p

22

p

2

2+ 2 p

=2p 2

2

2

2

p 2

s 2

p 2

2

s 2

p 2

2

p=

=

=

2

42

2

Then, when we rationalize this new denominator, we get that

pp p tan 22:5 = 2 p 2 p2 = 2 2

p 2=2 2

1p =2

1.

2

2

2

2

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

(c) cos 68 sin 8 sin 68 cos 8 = sin (8 68 ) = sin ( 60 ) = sin 60 = 3

2

2

tan + tan

(d)

15

5

2

1 tan

tan

2

23

5

p

= tan + = tan + = tan

= tan = 3

15 5

15 15

15

3

15

5

2. Prove each of the following identities.

cot2 x 1 (a) cot 2x =

2 cot x Solution:

cos2 x

cos2 x sin2 x cos2 x sin2 x

RHS

=

cot2 x 1 2 cot x

=

sin2 x cos x

2

1

=

sin2 x sin2 x cos x

2

=

sin2 x 2 cos x

=

sin x

sin x

sin x

cos2 x sin2 x sin x cos2 x sin2 x cos 2x

=

sin2 x

=

=

= cot 2x = LHS

2 cos x 2 sin x cos x sin 2x

(b) 4 sin4 x = 1 2 cos 2x + cos2 2x Solution:

RHS = 1 2 cos 2x + cos2 2x = 1 2 1 2 sin2 x + 1 2 sin2 x 2 = = 1 2 + 4 sin2 x + 1 4 sin2 x + 4 sin4 x = 4 sin4 x = LHS

(c) cos 3x = 4 cos3 x 3 cos x Solution:

LHS = 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) = = 2 cos3 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

cos2 x cos x = 2 cos x + 2 cos3 x =

2

2 cos 2x + 1 (d) tan 3x = tan x

2 cos 2x 1 Solution:

2 tan x

LHS = tan 3x = tan (x + 2x) = tan x + tan 2x =

tan x + 1

tan2 x

=

1 tan x tan 2x

2 tan x

1

tan x 1

tan2 x

2

tan x 1 + 1

tan2 x

=

= tan x

2

1+ 1

tan2 x

=

2 tan x

2 tan x

1

tan x 01

tan2 x

1

tan x 11

tan2 x

2

= tan x BB@

1+ 1

tan2 x

2 tan x

1 1

tan2 tan2

x x

CCA

=

tan

x

1

1 tan2 x + 2

tan2 x

= tan x (2 tan x)

1

tan x 1

tan2 x

3 tan2 x

3 tan2 x

= tan x 1

tan2 x

2 tan2 x = tan x 1

3 tan2 x =

3 = tan x

1

sin2 x

cos2 x sin2 x 3 cos2 x

cos2 x

3 cos2 x sin2 x

cos2 x = tan x cos2 x 3 sin2 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 2 cos2 x

sin2 x

cos2 x + sin2 x = tan x 2 cos 2x 1 = RHS

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 = 1 2 sin2 x 2 sin2 x + 3 sin x + 1 = 0 (sin x + 1) (2 sin x + 1) = 0

sin x + 1 = 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 sin x = 1 2

x= x=

+ 2k where k is an integer and 6 5 6 + 2k where k is an integer.

So the set of all solutions is

(b) sin 2x = 2 cos x Solution:

x= x= x=

+ 2k 2

+ 2k 6 5 6 + 2k where k is an integer.

sin 2x = 2 cos x 2 sin x cos x = 2 cos x 2 sin x cos x 2 cos x = 0 2 cos x (sin x 1) = 0

cos x = 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.

2p (c) cos 3x = 3

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

x= x=

50

+ k 120

where k is an integer

180

180

5 + k 2 where k is an integer.

18

3

8

12

4. Suppose that sin = and is not in the fourth quadrant; cos = and is not in the ...rst

17

13

quadrant. Find the exact value for each of the following.

Solution: The conditions given place into the third quadrant and into the fourth quadrant. These will determine the signs of the other trigonometric functions.

Since is in the third quadrant, cos is negative. For the rest,

s p cos = 1 sin2 = 1

82

15

=

17

17

Similarly, since

is in the fourth quadrant, sin is negative. For the rest,

q

s

sin = 1 cos 2 = 1

12 2

5

=

13

13

Now we have all we need:

8 sin =

17 cos = 15

17 tan = 8

15 5

sin = 13

12 cos = 13

5 tan = 12

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 = 22: 619 864 948 04 13

and . These will give us

(a) tan ( ) = Solution:

tan (

) = tan tan 1 + tan tan

8 = 15

1+ 8 15

5

8 (4) + 5 (5) 57

12 = 5 12

60 2

1 9

=

60 7

=

19 20

9 7

= 171 140

9

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download