Sample Midterm Exam - SOLUTIONS - Marta Hidegkuti

[Pages:7]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

171 We can use the calculator to check our result by entering the fraction = 1: 221 43 and

140 tan (208: 072 486 935 8 ( 22: 619 864 948 04 )) = 1: 221 43 and compare the decimals.

(b) cos ( + ) = Solution: cos ( + ) = cos cos

15 12 sin sin =

17 13

85

220

=

17 13

221

(c) sin 2 =

8 15

240

Solution: sin 2 = 2 sin = 2 17 17 = 289 = 0:830 449 826 989 6

(d)

sin 2 = Solution:

v u u t1 +

15 17

=

p4

= 0:970 142 500 145 3

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 25

y= x 33

and

5x + 3y = 1

3y = y=

5x + 1 51 3x + 3

2

5

So the slopes are m1 = tan = 3 and m2 = tan = 3 . We are now looking for tan (

).

52

tan (

tan tan )=

1 + tan tan

=

3

3 5

2

= 21

1+

33

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 sin 18 = 2 = x

5 10 x = 10 sin 18

To ...nd y:

cos 18 = y 5

y = 5 cos 18

Thus the triangle AOB has area

A = 1xy = 1 (10 sin 18 ) (5 cos 18 )

2

2

= 25 sin 18 cos 18

Thus the polygon's area is

10A = 10 (25 sin 18 cos 18 ) = 250 sin 18 cos 18 = = 73: 473 16 m2

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