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Lecture Notes
Simplifying Trigonometric Expressions
page 1
Sample Problems
Find the exact value of each of the following and simplify.
1. 2 sin 60
5 tan 30 cot 45
2. 2 sin 60
3 cos 45 + tan 30
3. 4 cos 30
sin 45 + tan 60
4. sin 30 cos 45 tan 60
5.
2 sec 60 +3 csc 45 tan 30 cos 60
sin 60
1
cos 30 + 1
Practice Problems
Find the exact value of each of the following and simplify.
1. sin 45
2. 3 tan 60
3.
2 cos 60 tan 30
csc 45
3 cos 45 sec 60
cos 45 tan2 60
5. 3 sec 30
cos 60 sin 60
tan 60
tan 30
tan 60 + tan 30
7.
sin 30 cos 30 tan 30 cot 30
sin 45 cos 45 tan 45 cot 45
tan 30 tan 60 csc 45
cos 45
cot 45
sin 45 + tan 45
4. sin2 30
6.
8.
2 cot 60 tan 60
c copyright Hidegkuti, Powell, 2012
9.
1
2 tan 30
tan2 30
(sin 30 ) (cos 30 )
(sin 45 ) (cos 45 )
(tan 30 ) cot 30
(tan 45 ) cot 45
Last revised: September 27, 2012
Simplifying Trigonometric Expressions
Lecture Notes
page 2
Sample Problems - Answers
p
2 3
3
1.)
3p
6
4.)
4
2.)
p
8 3
p
9 2
p
4 3
or
3
6
p
3 6 16
4 or
4
5.)
p
3 2
2
p
3.) 3 3
p
p
p
2
6 3
2
or
2
2
p
7+4 3
Practice Problems - Answers
p
2
2
1.)
8.)
p
3
p
3
3
9.) 2
p
2.) 3 3
p
4 2
p
3.) 2 2
3
4.)
p
3 2
2
7
4
7p
5.)
3
4
1
6.)
2
7.)
p
3
2
p
3
2
c copyright Hidegkuti, Powell, 2012
Last revised: September 27, 2012
Simplifying Trigonometric Expressions
Lecture Notes
page 3
Sample Problems - Solutions
Find the exact value of each of the following and simplify.
p
2 3
1. 2 sin 60
5 tan 30 cot 45 =
3
Solution: We …rst obtain the exact values of the trigonometric expressions. Using the famous right triangles,
we obtain that
p
3
1
sin 60 =
tan 30 = p
and
cot 45 = 1
2
3
Because we will be adding and subtracting these expressions, it is best if we rationalize them …rst.
p
p
3
3
1
tan 30 = p p =
3
3
3
p
p
p
p
3
3
5 3
5
1= 3
2 sin 60
5 tan 30 cot 45 = 2
2
3
3
We will then bring these to the common denominator
p
2. 2 sin 60
3
p
p
5 3
3 3
=
3
3
p
p
p
5 3
3 3 5 3
=
=
3
3
p
2 3
3
3 cos 45 + tan 30
Solution: We …rst obtain the exact values of the trigonometric expressions. Using the famous right triangles,
we obtain that
p
3
1
1
cos 45 = p
and
tan 30 = p
sin 60 =
2
2
3
Because we will be adding and subtracting these expressions, it is best if we rationalize them …rst.
p
p
p
p
3
3
2
1
2
1
and tan 30 = p p =
cos 45 = p p =
3
2
3
3
2
2
p
p
p
p
p
3
3 2
2
3 p
3
+
2 sin 60
3 cos 45 + tan 30 = 2
3
+
= 3
2
3
2
2
3
We will then bring these to the common denominator, which is 6 and if there are like terms, we combine them.
p
p
p
p
p
p
6 3 9 2 2 3
3
3 2
=
+
3
+
3
6
6
6
2
p
p
p
p
p
6 3 9 2+2 3
8 3 9 2
=
=
6
6
Note:
p
8 3
3. 4 cos 30
p
9 2
6
p
8 3
=
6
p
p
9 2
4 3
=
6
3
p
3 2
is also perfectly acceptable as …nal answer.
2
sin 45 + tan 60
Solution: We …rst obtain the exact values of the trigonometric expressions. Using the famous right triangles,
we obtain that
p
p
1
3
cos 30 =
sin 45 = p
and
tan 60 = 3
2
2
c copyright Hidegkuti, Powell, 2012
Last revised: September 27, 2012
Lecture Notes
Simplifying Trigonometric Expressions
page 4
Because we will be adding and subtracting these expressions, it is best if we rationalize them …rst.
p
p
1
2
2
sin 45 = p p =
2
2
2
p
p
p
p
p
p
3
2 p
2
2 p
4 cos 30
sin 45 + tan 60 = 4
+ 3=2 3
+ 3= 3 3
2
2
2
2
Note: We can also bring these to the common denominator.
p
p
p
p
p
p
2
6 3
2
6 3
2
3 3
=
=
2
2
2
2
is also perfectly acceptable.
3p
6 4
4
Solution: We …rst obtain the exact values of the trigonometric expressions. Using the famous right triangles,
we obtain that
p
p
1
1
1
1
sin 30 =
cos 45 = p tan 60 = 3 sec 60 = 2 csc 45 = 2 tan 30 = p and cos 60 =
2
2
3
2
4. sin 30 cos 45 tan 60
2 sec 60 + 3 csc 45 tan 30 cos 60 =
Because we will be adding and subtracting these expressions, it is best if we rationalize them …rst.
p
p
2
3
and tan 30 =
cos 45 =
2
3
sin 30 cos 45 tan 60
2 sec 60 + 3 csc 45 tan 30 cos 60 =
p
p
p
p
p
p
p
p
p
p
3 1
6
3 6
6
6
6
2 6
3 6
1
2 p
3 2 2+3
2
=
4+
=
4+
=
4+
=
=
6
4
2
4
4
4
3 2
4
2 2
Note: We can also bring these to the common denominator.
p
p
p
3 6
3 6 16
3 6 16
4=
=
4
4
4
4
4
is also perfectly acceptable.
p
sin 60
1
5.
= 7+4 3
cos 30 + 1
Solution: We …rst obtain the exact values of the trigonometric expressions. Using the famous right triangles,
we obtain that
p
3
p
p
1
sin 60
1
3
3
sin 60 =
and cos 30 =
and so
= p2
2
2
cos 30 + 1
3
+1
2
To clear the denominators, we multiply both numerator and denominator by 2
!
p
p
3
3
2
1
p
1
2
3 2
!=p
p2
=
p
3+2
3
3
+1
2
+1
2
2
We rationalize our result:
p
p
p
3 2
3 2
3
p
p
=p
3+2
3+2
3
2
=
2
p
3
3
2
4
2
=
3
p
4 3+4
7
=
1
p
4 3
=
1
7
p
4 3 =
p
7+4 3
For more documents like this, visit our page at http://www.teaching.martahidegkuti.com and click on Lecture
Notes. E-mail questions or comments to [email protected].
c copyright Hidegkuti, Powell, 2012
Last revised: September 27, 2012
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