Solution 1. Solution 2.

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

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