Section 4.3, Right Triangle Trigonometry
Homework: 4.3 #1–31 odds, 35, 37, 41
1
Another Approach for Calculating Trigonometric Functions
The techniques of this function work best when using acute angles, since we can draw any acute
angle as part of a right triangle.
Q
θ QQ hypotenuse
adjacent
Q
Q
Q
opposite
For the values of sine, cosine, and tangent, you can use “SOH-CAH-TOA.”
sin θ =
cos θ =
tan θ =
csc θ =
sec θ =
cot θ =
opposite
hypotenuse
adjacent
hypotenuse
opposite
=
adjacent
hypotenuse
opposite
hypotenuse
adjacent
adjacent
=
opposite
sin θ
cos θ
1
=
sin θ
1
=
cos θ
1
cos θ
=
tan θ
sin θ
Examples
1. Find all 6 trigonometric functions for θ.
θ
12
5
First, we need to calculate the length of the hypotenuse, which is
the Pythagorean Theorem.
opposite
hypotenuse
adjacent
cos θ =
hypotenuse
opposite
tan θ =
=
adjacent
sin θ =
5
13
12
=
13
5
12
=
√
122 + 52 =
hypotenuse
opposite
hypotenuse
sec θ =
adjacent
adjacent
cot θ =
=
opposite
csc θ =
13
5
13
=
12
12
5
=
√
169 = 13 by
2. Calculate all 6 trigonometric functions for π/6 (without using the unit circle). First, we will
draw the triangle in blue, then include the “mirror image” of the triangle in red.
Jπ
6J
J
J
J
J
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:
√
(The
√
Jπ
6J
3
J2
J
J
J
1
3 is from using the Pythagorean Theorem). Then,
π
1
=
6
2√
π
3
cos =
6
2
sin
√
1
π
3
tan = √ =
6
3
3
π
=2
6
√
π
2
2 3
sec = √ =
6
3
3
π √
cot = 3
6
csc
3. Calculate all 6 trigonometric functions for π/4 (without using the unit circle).
@
π
4
1
@
√
@ 2
@
@
1
@
This is an equilateral triangle, so the legs have the same length (here,
we’re assuming that
√
it’s 1). By the Pythagorean Theorem, the hypotenuse has length 2, so the 6 trigonometric
functions are:
√
π
1
2
π √
√
sin =
=
csc = 2
6
2
6
2
√
π
2
π √
cos =
sec = 2
6
2
6
π
π
tan = 1
cot = 1
6
6
Note: There is a table of values of sine, cosine, and tangent for “special” angles on page 303.
Memorize θ = π3 , π4 , π6 .
Example √
Let 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
More Trigonometric Properties and Identities
Cofunctions of complementary angles are equal. So, if θ is acute,
sin(90◦ − θ) = cos θ
tan(90◦ − θ) = cot θ
cos(90◦ − θ) = sin 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.
Example
Let 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 =
cos θ =
2
3
√
3
2
√
3
3 5
sec θ = √ =
5
5
√
5
cot θ =
2
csc θ =
5
3
√
2
2 5
tan θ = √ =
5
5
√
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 .)
Examples
Show each of the following identities:
1. (sec θ + 1)(sec θ − 1) = tan2 θ
When showing identities, it is normally easier to start with the more complicated side and
simplifying 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
= sin θ
sec θ
tan θ
=
sec θ
sin θ
cos θ
1
cos θ
= sin θ
3.
tan β + cot β
= csc2 β (#42)
tan β
tan β + cot β
= 1 + cot2 β
tan β
= csc2 β