Lesson Objectives
• Learn how to find the trigonometric function values of an acute angle
• Learn how to solve equations using cofunction identities
• Learn how to find the trigonometric function values of special angles

## How to Find the Trigonometric Function Values of an Acute Angle

In this lesson, we will learn about trigonometric functions of acute angles. Previously, we used angles in standard position to define the six trigonometric functions. Now, we will define our six trigonometric functions as ratios of the lengths of the sides of right triangles. Let's take a look at an acute angle in standard position: Notice the three points that are used to form our right triangle. The point (0,0) or the origin, the point (x,y), which can be any point on the terminal side that is not the origin, and the point (x,0). The vertical leg has a length of y since y - 0 = y and is known as the side opposite of our angle θ. The horizontal leg has a length of x since x - 0 = x and is known as the side adjacent to our angle θ. Additionally, we refer to the length of the hypotenuse, which extends from (0,0) to the point (x,y) as r. Using these new terms, let's revisit our definitions for the six trigonometric functions.

### Right-Triangle-Based Definitions of Trigonometric Functions

$$\text{sin}\hspace{.2em}θ=\frac{y}{r}=\frac{\text{opposite}}{\text{hypotenuse}}$$ $$\text{cos}\hspace{.2em}θ=\frac{x}{r}=\frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\text{tan}\hspace{.2em}θ=\frac{y}{x}=\frac{\text{opposite}}{\text{adjacent}}$$ $$\text{csc}\hspace{.2em}θ=\frac{r}{y}=\frac{\text{hypotenuse}}{\text{opposite}}$$ $$\text{sec}\hspace{.2em}θ=\frac{r}{x}=\frac{\text{hypotenuse}}{\text{adjacent}}$$ $$\text{cot}\hspace{.2em}θ=\frac{x}{y}=\frac{\text{adjacent}}{\text{opposite}}$$ Most textbooks will use: SOH - CAH - TOA as a way to remember the three main trigonometric functions. The other three can be found using the reciprocal identities.
SOH... » Sine = Opposite/Hypotenuse
Here, the first letter S is for Sine (sin), the next two letters give the ratio: opposite/hypotenuse.
Here, the first letter C is for Cosine (cos), the next two letters give the ratio: adjacent/hypotenuse.
Lastly, the first letter T is for Tangent (tan), the next two letters give the ratio: opposite/adjacent. Let's look at an example.
Example #1: Find all six trigonometric functions given the lengths of a, b, and c. $$a=2$$ $$b=4$$ $$c=2\sqrt{5}$$ Our side opposite of our angle θ is labeled as "a". Our side that is adjacent to our angle θ is labeled as "b". Our hypotenuse is labeled as "c". Let's plug in the values given for the lengths of a, b, and c into our formulas: $$\text{sin}\hspace{.2em}θ=\frac{\text{opposite}}{\text{hypotenuse}}$$ $$\text{sin}\hspace{.2em}θ=\frac{2}{2\sqrt{5}}=\frac{\sqrt{5}}{5}$$ $$\text{csc}\hspace{.2em}θ=\frac{\text{hypotenuse}}{\text{opposite}}$$ $$\text{csc}\hspace{.2em}θ=\frac{2\sqrt{5}}{2}=\sqrt{5}$$
$$\text{cos}\hspace{.2em}θ=\frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\text{cos}\hspace{.2em}θ=\frac{4}{2\sqrt{5}}=\frac{2\sqrt{5}}{5}$$ $$\text{sec}\hspace{.2em}θ=\frac{\text{hypotenuse}}{\text{adjacent}}$$ $$\text{sec}\hspace{.2em}θ=\frac{2\sqrt{5}}{4}=\frac{\sqrt{5}}{2}$$
$$\text{tan}\hspace{.2em}θ=\frac{\text{opposite}}{\text{adjacent}}$$ $$\text{tan}\hspace{.2em}θ=\frac{2}{4}=\frac{1}{2}$$ $$\text{cot}\hspace{.2em}θ=\frac{x}{y}=\frac{\text{adjacent}}{\text{opposite}}$$ $$\text{cot}\hspace{.2em}θ=\frac{4}{2}=2$$ In order to understand our next topic, let's use the same triangle but change our angle θ.
Example #2: Find all six trigonometric functions given the lengths of a, b, and c. $$a=2$$ $$b=4$$ $$c=2\sqrt{5}$$ Our side opposite of our angle θ is now labeled as "b". Our side that is adjacent to our angle θ is now labeled as "a". Our hypotenuse is still labeled as "c". Let's plug in the values given for the lengths of a, b, and c into our formula: $$\text{sin}\hspace{.2em}θ=\frac{\text{opposite}}{\text{hypotenuse}}$$ $$\text{sin}\hspace{.2em}θ=\frac{4}{2\sqrt{5}}=\frac{2\sqrt{5}}{5}$$ $$\text{csc}\hspace{.2em}θ=\frac{\text{hypotenuse}}{\text{opposite}}$$ $$\text{csc}\hspace{.2em}θ=\frac{2\sqrt{5}}{4}=\frac{\sqrt{5}}{2}$$
$$\text{cos}\hspace{.2em}θ=\frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\text{cos}\hspace{.2em}θ=\frac{2}{2\sqrt{5}}=\frac{\sqrt{5}}{5}$$ $$\text{sec}\hspace{.2em}θ=\frac{\text{hypotenuse}}{\text{adjacent}}$$ $$\text{sec}\hspace{.2em}θ=\frac{2\sqrt{5}}{2}=\sqrt{5}$$
$$\text{tan}\hspace{.2em}θ=\frac{\text{opposite}}{\text{adjacent}}$$ $$\text{tan}\hspace{.2em}θ=\frac{4}{2}=2$$ $$\text{cot}\hspace{.2em}θ=\frac{\text{adjacent}}{\text{opposite}}$$ $$\text{cot}\hspace{.2em}θ=\frac{2}{4}=\frac{1}{2}$$

### Cofunction Identities

Notice that the sin θ from the first example is equal to the cos θ from the second example. Additionally, you can see that cos θ from the first example is equal to sin θ from the second example. Let's look at our same right triangle with two acute angles A and B: $$\text{sin}\hspace{.2em}A=\frac{a}{c}=\text{cos}\hspace{.2em}B$$ $$\text{tan}\hspace{.2em}A=\frac{a}{b}=\text{cot}\hspace{.2em}B$$ $$\text{sec}\hspace{.2em}A=\frac{c}{b}=\text{csc}\hspace{.2em}B$$ These relationships are always true for the two acute angles of a right triangle. Since the sum of the three angles in any triangle is 180° and angle C is exactly 90°. This means that angles A and B will have a sum of 90°. We know that two angles whose sum is 90° are known as complementary angles. Since our angles A and B are complementary and sin A = cos B, we can say the functions sine and cosine are known as cofunctions. We can also say that tangent and cotangent are cofunctions, and secant and cosecant. For any acute angle A, cofunction values of complementary angles are equal:
Cofunction Identities
sin A = cos(90° - A)
cos A = sin(90° - A)
sec A = csc(90° - A)
csc A = sec(90° - A)
tan A = cot(90° - A)
cot A = tan(90° - A)
Let's look at an example.
Example #3: Write each function in terms of its cofunction.
cos 48°
cos 48° = sin(90° - 48°) = sin 42°
Example #4: Write each function in terms of its cofunction.
tan 11°
tan 11° = cot(90° - 11°) = cot 79°

### Solving Equations Using Cofunction Identities

Let's look at an example that involves solving equations using cofunction identities.
Example #5: Find one solution for each equation. Assume all angles involved are acute angles.
sin(2θ + 20°) = cos(3θ + 5°)
We know that sine and cosine are cofunctions:
sin(2θ + 20°) = cos(3θ + 5°) is only true if the sum of the angles is 90°
2θ + 20° + 3θ + 5° = 90°
Let's combine like terms:
5θ + 25° = 90°
Isolate θ
5θ = 90° - 25°
5θ = 65°
θ = (65/5)°
θ = 13°
Check:
sin(2(13°) + 20°) = cos(3(13°) + 5°)
sin(46°) = cos(44°) ✓

### Trigonometric Function Values of Special Angles

Certain angles appear very frequently: 30°, 45°, and 60°. The function values of these special angles can be summarized using the following table:
θ sin θ cos θ tan θ
30°$\frac{1}{2}$$\frac{\sqrt{3}}{2}$$\frac{\sqrt{3}}{3}$
45°$\frac{\sqrt{2}}{2}$$\frac{\sqrt{2}}{2}$$1$
60°$\frac{\sqrt{3}}{2}$$\frac{1}{2}$$\sqrt{3}$
The other three function values can be found using the reciprocal identities:
θ cot θ sec θ csc θ
30°$\sqrt{3}$$\frac{2\sqrt{3}}{3}$$2$
45°$1$$\sqrt{2}$$\sqrt{2}$
60°$\frac{\sqrt{3}}{3}$$2$$\frac{2\sqrt{3}}{3}$

#### Skills Check:

Example #1

Find csc θ

A
$$\text{csc}\hspace{.15em}θ=\frac{\sqrt{15}}{7}$$
B
$$\text{csc}\hspace{.15em}θ=\frac{\sqrt{15}}{8}$$
C
$$\text{csc}\hspace{.15em}θ=\frac{2\sqrt{6}}{7}$$
D
$$\text{csc}\hspace{.15em}θ=\frac{7}{8}$$
E
$$\text{csc}\hspace{.15em}θ=\frac{8}{7}$$

Example #2

Write each function in terms of its cofunction.

sec 34°

A
sec 34° = csc 146°
B
sec 34° = cos 56°
C
sec 34° = csc 56°
D
sec 34° = cos 146°
E
sec 34° = tan 56°

Example #3

Find one solution for the equation. Assume all angles are acute angles.

tan(2θ - 25°) = cot(60° - θ)

A
θ = 15°
B
θ = 55°
C
θ = 25°
D
θ = 35°
E
θ = 20°