Using Right Triangles to Evaluate Trigonometric Functions

In earlier sections, we used a unit circle to define the trigonometric functions. In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of [latex]t[/latex] is its value at [latex]t[/latex] radians. First, we need to create our right triangle. Figure 1 shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point [latex]\left(x,y\right)\\[/latex] to the x-axis, we have a right triangle whose vertical side has length [latex]y[/latex] and whose horizontal side has length [latex]x[/latex]. We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.

Graph of quarter circle with radius of 1 and angle of t. Point of (x,y) is at intersection of terminal side of angle and edge of circle.

Figure 1

 

We know

[latex]\cos \text{ }t=\frac{x}{1}=x\\[/latex]

Likewise, we know

[latex]\sin \text{ }t=\frac{y}{1}=y\\[/latex]

These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using [latex]\left(x,y\right)\\[/latex] coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of [latex]x[/latex], we will call the side between the given angle and the right angle the adjacent side to angle [latex]t[/latex]. (Adjacent means “next to.”) Instead of [latex]y[/latex], we will call the side most distant from the given angle the opposite side from angle [latex]\text{}t[/latex]. And instead of [latex]1[/latex], we will call the side of a right triangle opposite the right angle the hypotenuse. These sides are labeled in Figure 2.

A right triangle with hypotenuse, opposite, and adjacent sides labeled.

Figure 2. The sides of a right triangle in relation to angle [latex]t[/latex].

Understanding Right Triangle Relationships

Given a right triangle with an acute angle of [latex]t[/latex],

[latex]\begin{array}{l}\sin \left(t\right)=\frac{\text{opposite}}{\text{hypotenuse}}\hfill \\ \cos \left(t\right)=\frac{\text{adjacent}}{\text{hypotenuse}}\hfill \\ \tan \left(t\right)=\frac{\text{opposite}}{\text{adjacent}}\hfill \end{array}\\[/latex]

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.”

How To: Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle.

  1. Find the sine as the ratio of the opposite side to the hypotenuse.
  2. Find the cosine as the ratio of the adjacent side to the hypotenuse.
  3. Find the tangent is the ratio of the opposite side to the adjacent side.

Example 1: Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure 3, find the value of [latex]\cos \alpha \\[/latex].

A right triangle with sid lengths of 8, 15, and 17. Angle alpha also labeled.

Figure 3

 

Solution

The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so:

[latex]\begin{array}{l}\cos \left(\alpha \right)=\frac{\text{adjacent}}{\text{hypotenuse}}\hfill \\ =\frac{15}{17}\hfill \end{array}\\[/latex]

Try It 1

Given the triangle shown in Figure 4, find the value of [latex]\text{sin}t[/latex].

A right triangle with sides of 7, 24, and 25. Also labeled is angle t.

Figure 4

Solution

Relating Angles and Their Functions

When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure 5. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

Right triangle with angles alpha and beta. Sides are labeled hypotenuse, adjacent to alpha/opposite to beta, and adjacent to beta/opposite alpha.

Figure 5. The side adjacent to one angle is opposite the other.

We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

How To: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.

  1. If needed, draw the right triangle and label the angle provided.
  2. Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
  3. Find the required function:
    • sine as the ratio of the opposite side to the hypotenuse
    • cosine as the ratio of the adjacent side to the hypotenuse
    • tangent as the ratio of the opposite side to the adjacent side
    • secant as the ratio of the hypotenuse to the adjacent side
    • cosecant as the ratio of the hypotenuse to the opposite side
    • cotangent as the ratio of the adjacent side to the opposite side

Example 2: Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure 6, evaluate [latex]\sin \alpha \\[/latex], [latex]\cos \alpha \\[/latex], [latex]\tan \alpha \\[/latex], [latex]\sec \alpha\\ [/latex], [latex]\csc \alpha \\[/latex], and [latex]\cot \alpha\\ [/latex].

Right triangle with sides of 3, 4, and 5. Angle alpha is also labeled.

Figure 6

 

Solution

[latex]\begin{array}{l}\sin \alpha =\frac{\text{opposite }\alpha }{\text{hypotenuse}}=\frac{4}{5}\hfill \\ \cos \alpha =\frac{\text{adjacent to }\alpha }{\text{hypotenuse}}=\frac{3}{5}\hfill \\ \tan \alpha =\frac{\text{opposite }\alpha }{\text{adjacent to }\alpha }=\frac{4}{3}\hfill \\ \sec \alpha =\frac{\text{hypotenuse}}{\text{adjacent to }\alpha }=\frac{5}{3}\hfill \\ \csc \alpha =\frac{\text{hypotenuse}}{\text{opposite }\alpha }=\frac{5}{4}\hfill \\ \cot \alpha =\frac{\text{adjacent to }\alpha }{\text{opposite }\alpha }=\frac{3}{4}\hfill \end{array}\\[/latex]

Try It 2

Using the triangle shown in Figure 7, evaluate [latex]\sin \text{ }t\\[/latex], [latex]\cos \text{ }t\\[/latex], [latex]\tan \text{ }t\\[/latex], [latex]\sec \text{ }t\\[/latex], [latex]\csc \text{ }t\\[/latex], and [latex]\cot \text{ }t\\[/latex].

Right triangle with sides 33, 56, and 65. Angle t is also labeled.

Figure 7

Solution

Finding Trigonometric Functions of Special Angles Using Side Lengths

We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of [latex]30^\circ [/latex], [latex]60^\circ [/latex], and [latex]45^\circ \\[/latex], however, remember that when dealing with right triangles, we are limited to angles between [latex]0^\circ \text{ and 90^\circ }\text{.}\\[/latex]

Suppose we have a [latex]30^\circ ,60^\circ ,90^\circ \\[/latex] triangle, which can also be described as a [latex]\frac{\pi }{6},\text{ } \frac{\pi }{3},\frac{\pi }{2}\\[/latex] triangle. The sides have lengths in the relation [latex]s,\sqrt{3}s,2s\\[/latex]. The sides of a [latex]45^\circ ,45^\circ ,90^\circ [/latex] triangle, which can also be described as a [latex]\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}\\[/latex] triangle, have lengths in the relation [latex]s,s,\sqrt{2}s\\[/latex]. These relations are shown in Figure 8.

Two side by side graphs of circles with inscribed angles. First circle has angle of pi/3 inscribed. Second circle has angle of pi/4 inscribed.

Figure 8. Side lengths of special triangles

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

How To: Given trigonometric functions of a special angle, evaluate using side lengths.

  1. Use the side lengths shown in Figure 8 for the special angle you wish to evaluate.
  2. Use the ratio of side lengths appropriate to the function you wish to evaluate.

Example 3: Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of [latex]\frac{\pi }{3}\\[/latex], using side lengths.

Solution

[latex]\begin{array}{l}\sin \left(\frac{\pi }{3}\right)=\frac{\text{opp}}{\text{hyp}}=\frac{\sqrt{3}s}{2s}=\frac{\sqrt{3}}{2}\hfill \\ \cos \left(\frac{\pi }{3}\right)=\frac{\text{adj}}{\text{hyp}}=\frac{s}{2s}=\frac{1}{2}\hfill \\ \tan \left(\frac{\pi }{3}\right)=\frac{\text{opp}}{\text{adj}}=\frac{\sqrt{3}s}{s}=\sqrt{3}\hfill \\ \sec \left(\frac{\pi }{3}\right)=\frac{\text{hyp}}{\text{adj}}=\frac{2s}{s}=2\hfill \\ \csc \left(\frac{\pi }{3}\right)=\frac{\text{hyp}}{\text{opp}}=\frac{2s}{\sqrt{3}s}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\hfill \\ \cot \left(\frac{\pi }{3}\right)=\frac{\text{adj}}{\text{opp}}=\frac{s}{\sqrt{3}s}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\hfill \end{array}\\[/latex]

Try It 3

Find the exact value of the trigonometric functions of [latex]\frac{\pi }{4}\\[/latex], using side lengths.

Solution