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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