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