section 4.3, right triangle trigonometry - university of...
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Section 4.3, Right Triangle Trigonometry
Homework: 4.3 #1–31 odds, 35, 37, 41
1 Another Approach for Calculating Trigonometric Func-tions
The techniques of this function work best when using acute angles, since we can draw any acuteangle as part of a right triangle.
QQQQQQ
adjacent
opposite
θ hypotenuse
For the values of sine, cosine, and tangent, you can use “SOH-CAH-TOA.”
sin θ =opposite
hypotenuse
cos θ =adjacent
hypotenuse
tan θ =opposite
adjacent=
sin θ
cos θ
csc θ =hypotenuse
opposite=
1
sin θ
sec θ =hypotenuse
adjacent=
1
cos θ
cot θ =adjacent
opposite=
1
tan θ=
cos θ
sin θ
Examples
1. Find all 6 trigonometric functions for θ.
������
5
12θ
First, we need to calculate the length of the hypotenuse, which is√
122 + 52 =√
169 = 13 bythe Pythagorean Theorem.
sin θ =opposite
hypotenuse=
5
13csc θ =
hypotenuse
opposite=
13
5
cos θ =adjacent
hypotenuse=
12
13sec θ =
hypotenuse
adjacent=
13
12
tan θ =opposite
adjacent=
5
12cot θ =
adjacent
opposite=
12
5
1
2. Calculate all 6 trigonometric functions for π/6 (without using the unit circle). First, we willdraw the triangle in blue, then include the “mirror image” of the triangle in red.
JJJJJJJ
π6
These two triangles combined give us an equilateral triangle (since all 3 angles are equal.Specifically, they all measure π/6 = 60◦.). Let’s assume that every side has length 2. Then,we can label the (smaller) triangle:
JJJJJJJ
π6√
3
1
2
(The√
3 is from using the Pythagorean Theorem). Then,
sinπ
6=
1
2csc
π
6= 2
cosπ
6=
√3
2sec
π
6=
2√3
=2√
3
3
tanπ
6=
1√3
=
√3
3cot
π
6=√
3
3. Calculate all 6 trigonometric functions for π/4 (without using the unit circle).
@@@@@@
π4
1
1
√2
This is an equilateral triangle, so the legs have the same length (here, we’re assuming thatit’s 1). By the Pythagorean Theorem, the hypotenuse has length
√2, so the 6 trigonometric
functions are:
sinπ
6=
1√2
=
√2
2csc
π
6=√
2
cosπ
6=
√2
2sec
π
6=√
2
tanπ
6= 1 cot
π
6= 1
Note: There is a table of values of sine, cosine, and tangent for “special” angles on page 303.Memorize θ = π
3 ,π4 ,
π6 .
ExampleLet cot θ =
√3 for 0 ≤ θ ≤ π/2. Find θ in radians.
By the table on page 303, we know that θ = π/6.
2
2 More Trigonometric Properties and Identities
Cofunctions of complementary angles are equal. So, if θ is acute,
sin(90◦ − θ) = cos θ cos(90◦ − θ) = sin θ
tan(90◦ − θ) = cot θ cot(90◦ − θ) = tan θ
sec(90◦ − θ) = csc θ csc(90◦ − θ) = sec θ
90◦ can be replaced by π/2 if the angles are in radians.
ExampleLet sin θ = 2/3.
1. Calculate cos(π2 − θ).
cos(π2 − θ) = sin θ = 2/3
2. Calculate all 6 trig functions for θ.
sin θ =2
3csc θ =
3
2
cos θ =
√5
3sec θ =
3√5
=3√
5
5
tan θ =2√5
=2√
5
5cot θ =
√5
2
The√
5 is from the Pythagorean Theorem.
The Pythagorean Identities are:
sin2 θ + cos2 θ = 1 Dividing both sides by cos2 θ, we get that:
tan2 θ + 1 = sec2 θ Dividing both sides of the first line by sin2 θ, we get that:
1 + cot2 θ = csc2 θ
(Note: sin2 θ = (sin θ)2.)
ExamplesShow each of the following identities:
1. (sec θ + 1)(sec θ − 1) = tan2 θWhen showing identities, it is normally easier to start with the more complicated side andsimplifying it. Also note that there is more than one way to get to the correct answer.
(sec θ + 1)(sec θ − 1) = sec2 θ − 1
= tan2 θ
2.tan θ
sec θ= sin θ
tan θ
sec θ=
sin θcos θ1
cos θ
= sin θ
3.tanβ + cotβ
tanβ= csc2 β (#42)
tanβ + cotβ
tanβ= 1 + cot2 β
= csc2 β
3