Solution 1. Solution 2.

[Pages:14]Arkansas Tech University MATH 1203: Trigonometry

Dr. Marcel B. Finan

Solutions to Review Problems for Final Exam

Solution 1. Using the identity 1 + tan2 = sec2 we find

sec2 - 1 1 + tan2 - 1

=

sec2

sec2

tan2 = sec2

sin2 = cos2

cos2

=

sin2

Solution 2. Taking common denominator and using the identity cos2 + sin2 = 1 we

find

sin 1 + cos (1 + cos )2 + sin2

+

=

1 + cos sin

sin (1 + cos )

2(1 + cos ) =

sin (1 + cos )

=2 csc

Solution 3. Multiplying we find

(sin x - cos x)(sin x + cos x) = sin2 x - cos2 x

Solution 4.

1

Using

the quotient

identity tan x

=

sin x cos x

and the Pythagorean

identity cos2 x+

sin2 x = 1 we find

sin x

cos x + tan x sin x = cos x +

sin x

cos x

cos2 x + sin2 x =

cos x

1 = = sec x.

cos x

Solution 5. Using the identity cos2 + sin2 = 1 we have

sin2 1 - cos2 =

1 - cos 1 - cos (1 - cos )(1 + cos )

= 1 - cos

=1 + cos = cos (1 + sec )

1 + sec =

sec

Solution 6.

Letting

=

2

we

get

1

=

sin

2

=

cos

2

= 0.

Solution 7. The left-hand side looks more complex then the right-hand side, so we start with it and try to transform it to the right-hand side.

cos x(sec x - cos x) = cos x sec x - cos2 x

1 = cos x

= cos2 x

cos x

=1 - cos2 x = sin2 x

Solution 8.

2

Starting from the right-hand side to obtain

1

1

(1 + sin x) - (1 - sin x) 2 sin x

-

=

1 - sin x 1 + sin x (1 - sin x)(1 + sin x)

= 1 - sin2 x

2 sin x sin x 1

=

=2

= 2 tan x sec x

cos2 x cos x cos x

Solution 9. Using the conjugate of 1 - sin x to obtain

cos x

cos x(1 + sin x)

=

1 - sin x (1 - sin x)(1 + sin x)

cos x + cos x sin x = 1 - sin2 x

cos x + cos x sin x

=

cos2 x

1 sin x

=+

= sec x + tan x.

cos x cos x

Solution 10. Notice first that 75 = 30 + 45. Thus,

sin 75 = sin (45 + 30)

= sin 45 cos 30 + cos 45 sin 30

2 3 2 1 6+ 2

= ? + ?=

2 2 22

4

Solution 11.

3

Since

12

=

4

-

3

,

the

difference

formula

for

sine

gives

sin = sin ( - )

12

46

= sin cos - cos sin

4 6 4 6

2 3 2 1 6- 2

= ? - ?=

2 2 22

4

Solution 12. Since the sine function is an odd function, we can write

sin x = - sin (-x) = - sin [( - x) - ]

2

2

= - [sin ( - x) cos ( ) - cos ( - x) sin ( )]

2

2

2

2

= cos ( - x)

2

Solution 13. We have

7

cos = cos ( + )

12

43

= cos cos - sin sin

4 3 4 3

2 1 2 3 2- 6

= ?- ? =

22 2 2

4

Solution 14.

sin 42

cos 12

-

cos 42

sin 12

=

sin (42

-

12)

=

sin 30

=

1 2

Solution 15.

Since and are in the third quadrant, we have cos = -

1

-

sin2

=

-

1 2

and cos = -

1

-

sin2

=

-

3 2

.

Thus,

cos ( + ) =

cos cos - sin sin

=

(-

1 2

)(-

3 2

)

-

(-

3 2

)(-

1 2

)

=

0

Solution 16.

4

We

have

tan (

+

)

=

tan +tan 1-tan tan

=

tan

since

tan

=

0.

Solution 17.

The fact is in quadrant IV implies sin = -

1

-

cos2

=

-

12 13

.

Thus,

120 sin 2 =2 sin cos = -

169

cos

2

=2

cos2

-

1

=

119 -

169

sin 2 120

tan 2 =

=

cos 2 119

Solution 18. Using the formula for tan 2 we have

1

1 - tan2

cot 2 =

=

tan (2) 2 tan

11

1

= ( - tan ) = (cot - tan )

2 tan

2

Solution 19. We have

sin4

=(sin2

)2

=

1 (

-

cos

2

)2

2

= 1 (1 + cos2 2 - 2 cos 2) 4

1

1 + cos 4

= (1 + (

) - 2 cos 2)

4

2

31

1

= - cos 2 + cos 4

82

8

Solution 20.

5

Since is in quadrant II, we have cos = -

1

-

sin2

=

-

4 5

.

Thus,

sin =

2 =

1 - cos

2

1+

4 5

=

3 10

2

10

cos = -

2 =-

1 + cos

2

1-

4 5

10

=-

2

10

1 - cos

tan = -

= -3

(1)

2

1 + cos

Solution 21. Using ( ??) we obtain

1 sin 3x cos x = [sin (3x + x) + sin (3x - x)]

2 1 = (sin 4x + sin 2x) 2

Solution 22. Using the product-to-sum identities we find

cos 2x + cos 2y cos 2x - cos 2y

=

2

cos

(

2x+2y 2

)

cos

(

2x-2y 2

)

-2

sin

(

2x+2y 2

)

sin

(

2x-2y 2

)

= - cot (x + y) cot (x - y)

cos 2x + cos 2y cos 2x - cos 2y

=

2

cos

(

2x+2y 2

)

cos

(

2x-2y 2

)

-2

sin

(

2x+2y 2

)

sin

(

2x-2y 2

)

= - cot (x + y) cot (x - y)

Solution 23. 6

Since

a

=

1 2

and b

=

-

1 2

we

find

k

=

(

1 2

)2

+

(-

1 2

)2

=

2 2

,

cos

=

a k

=

2 2

,

sin

=

b k

=

-

2 2

.

Thus

=

-45

and

y = 2 sin (x - 45).

2

Solution 24.

(a)

Since

sin

2

=

1

we

find

sin-1 (1)

=2 .

(b) Since sin

3

=

3 2

we

find

sin-1

3 2

=

3

.

(c) Since sin

-

6

=

-

1 2

we

find

sin-1

-

1 2

=

-

6

.

Solution 25.

(a) sin (sin-1 2) is undefined since 2 is not in the domain of sin-1 x.

(b)

sin (sin-1

3

)

=

3

.

Solution 26.

(a) Using the above discussion we find cos (sin-1

2 2

)

=

-1

(b)

tan

(sin-1

(-

1 2

))

=

2

1-

1 4

=-

3 3

.

Solution 27.

(a) cos-1

2 2

=

4

since

cos

4

=

2 2

.

(b) cos-1

-

1 2

=

2 3

.

Solution 28.

(a)

tan-1 (tan

4

)

=

4

.

(b)

tan-1 (tan

7 5

)

=

tan-1

(tan

(

2 5

))

=

2 5

.

1

-

(

2 2

)2

=

2 2

.

Solution 29.

The

given

equation

is

equivalent

to

sin x

=

1 2

=

sin

6

.

The

solutions

to

this

equation are given by

x

=

6

+

2k

and

x

=

5 6

+

2k

Solution 30.

Since sin x = sin

sin-1

1 3

, the solutions are given by

x = sin-1

1 3

+ 2k and x = - sin-1

1 3

+ 2k

7

Solution 31.

Factoring we find sin x(sin x - 1) = 0. Thus, either sin x = 0 or sin x = 1.

The solutions of the equation sin x = 0 are given by x = k where k is any

integer.

The

solutions

of

the

equation

sin x = 1

are

given

by

x

=

(2k

+

1)

2

where k is an arbitrary integer.

Solution 32. Factoring the given equation to obtain:

(2 cos x - 1)(cos x - 3) = 0.

This

equation

is

satisfied

for

all

values

of

x

such

that

either

cos x

=

1 2

or

cos x = 3. Since -1 cos x 1, the second equation has no solutions. The

solutions

to

the

first

equation

are

given

by

3

+ 2k

or

-

3

+ 2k

where

k

is

an integer.

Solution 33.

Using the identity sin2 x + cos2 x = 1 we obtain the quadratic equation

2 cos2 x+3 cos x+1 = 0 which can be factored into (2 cos x+1)(cos x+1) = 0.

Thus

either

cos x

=

-

1 2

or

cos x

=

-1.

The

solutions

to

the

first

equation

are

given by

x

=

2 3

+ 2k

and

x

=

-

2 3

+ 2k

The solutions to the second equation are given by x = (2k + 1) where k is

an arbitrary integer.

Solution 34.

Using the identity sin 2x = 2 sin x cos x the given equation can be factored

as

cos x(2 sin x - 1)

=

0.

Thus,

either

cos x

=

0

or

sin x

=

1 2

.

The

solutions

to

the

first

equation

are

given

by

x

=

(2k

+

1)

2

and

those

to

the

second

equation are given by

x

=

6

+

2k

and

x

=

-

6

+

2k

where k is an integer

Solution 35. Squaring both sides of the equation and expanding to obtain

cos2 x + 2 cos x + 1 = sin2 x

8

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