3.1: The Trigonometric Functions - Right Triangle Definition

Learning Objectives Find the value of a trigonometric function of an unknown angle in a right triangle given at least two sides of the triangle. Compute the values of the remaining trigonometric functions of an unknown angle within a right triangle given one trigonometric ratio. Use the Ratio, Reciprocal, and Pythagorean Identities, along with Right Triangle Trigonometry, to compute trigonometric values. Find the exact value of a trigonometric function for a special angle. Use the Cofunction Identities to find the value of a trigonometric function. Use exact values to simplify an expression involving trigonometric functions. Recognize and understand the constraints on the values of certain trigonometric functions.

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Introduction to Right Triangle Trigonometry

Our initial foray into Trigonometry involved defining the trigonometric functions in terms of points on the Cartesian plane and the distances from those points to the origin. In this chapter, we focus on Trigonometry outside of a coordinate system. Specifically, we redefine the trigonometric functions in terms of ratios of sides of a right triangle.

Definition: Trigonometric Functions (Right Triangle Definition)

If $\theta$ is one of the angles in a right triangle, where the side opposite $\theta$ is called "opposite," the side adjacent to $\theta$ is called "adjacent," and the hypotenuse is named as such, we define the trigonometric functions of $\theta$ as follows: $$\begin{array}{|ccc|ccc|} \hline \text{Function} & & \text{Ratio} & \text{Function} & & \text{Ratio} \\[6pt] \hline \sin\left( \theta \right) & = & \dfrac{\text{opposite}}{\text{hypotenuse}} & \csc\left( \theta \right) & = & \dfrac{\text{hypotenuse}}{\text{opposite}} \\[6pt] \cos\left( \theta \right) & = & \dfrac{\text{adjacent}}{\text{hypotenuse}} & \sec\left( \theta \right) & = & \dfrac{\text{hypotenuse}}{\text{adjacent}} \\[6pt] \tan\left( \theta \right) & = & \dfrac{\text{opposite}}{\text{adjacent}} & \cot\left( \theta \right) & = & \dfrac{\text{adjacent}}{\text{opposite}} \\[6pt] \hline \end{array} \nonumber$$

While this definition looks different than our coordinate definition of the trigonometric functions, it is, in fact, the same definition if we restrict the coordinates in our original definition to the first quadrant. While it might seem limiting for us to create a new definition of the trigonometric functions by restricting our original definition, this new definition serves a greater purpose - it simplifies Trigonometry immensely. It allows us to separate the triangle from the Cartesian coordinate system.

Example $\PageIndex{ 1 }$

Find all trigonometric ratios of the labeled angle in each triangle.

1.  

   Figure $\PageIndex{ 1 }$

   

2.  

   Figure $\PageIndex{ 2 }$

   

Solutions

1. The side opposite angle $\phi$ is 5 and the hypotenuse is 13. Let $x$ be the length of the side adjacent to $\phi$. To find $x$, we invoke the Pythagorean Theorem. $$\begin{array}{rrclcr} & x^2 + 5^2 & = & 13^2 & \quad & (\text{Pythagorean Theorem}) \\[6pt] \implies & x^2 + 25 & = & 169 & \quad & (\text{simplifying}) \\[6pt] \implies & x^2 & = & 144 & \quad & (\text{subtracting } 25\text{ from both sides}) \\[6pt] \implies & x & = & \pm 12 & \quad & (\text{Extraction of Roots}) \\[6pt] \implies & x & = & 12 & \quad & (x\text{ must be positive - it's a side length}) \\[6pt] \end{array} \nonumber$$ Therefore, $$\begin{array}{rccclcrcccl} \sin\left( \phi \right) & = & \dfrac{\text{opposite}}{\text{hypotenuse}} & = & \dfrac{5}{13} & \qquad & \csc\left( \phi \right) & = & \dfrac{\text{hypotenuse}}{\text{opposite}} & = & \dfrac{13}{5} \\[6pt] \cos\left( \phi \right) & = & \dfrac{\text{adjacent}}{\text{hypotenuse}} & = & \dfrac{12}{13} & \qquad & \sec\left( \phi \right) & = & \dfrac{\text{hypotenuse}}{\text{adjacent}} & = & \dfrac{13}{12} \\[6pt] \tan\left( \phi \right) & = & \dfrac{\text{opposite}}{\text{adjacent}} & = & \dfrac{5}{12} & \qquad & \cot\left( \phi \right) & = & \dfrac{\text{adjacent}}{\text{opposite}} & = & \dfrac{12}{5} \\[6pt] \end{array} \nonumber$$
2. The side opposite angle $\beta$ is $\sqrt{5}$ and the hypotenuse is 3. Labeling the side adjacent to $\beta$ as $x$, we get $$\begin{array}{rrclcr} & x^2 + \left(\sqrt{5}\right)^2 & = & 3^2 & \quad & (\text{Pythagorean Theorem}) \\[6pt] \implies & x^2 + 5 & = & 9 & \quad & (\text{simplifying}) \\[6pt] \implies & x^2 & = & 4 & \quad & (\text{subtracting } 5 \text{ from both sides}) \\[6pt] \implies & x & = & 2 & \quad & (\text{Extraction of Roots}) \\[6pt] \end{array} \nonumber$$ Therefore, $$\begin{array}{rccclcrcccl} \sin\left( \beta \right) & = & \dfrac{\text{opposite}}{\text{hypotenuse}} & = & \dfrac{\sqrt{5}}{3} & \qquad & \csc\left( \beta \right) & = & \dfrac{\text{hypotenuse}}{\text{opposite}} & = & \dfrac{\sqrt{3}}{\sqrt{5}} \\[6pt] \cos\left( \beta \right) & = & \dfrac{\text{adjacent}}{\text{hypotenuse}} & = & \dfrac{2}{3} & \qquad & \sec\left( \beta \right) & = & \dfrac{\text{hypotenuse}}{\text{adjacent}} & = & \dfrac{3}{2} \\[6pt] \tan\left( \beta \right) & = & \dfrac{\text{opposite}}{\text{adjacent}} & = & \dfrac{\sqrt{5}}{2} & \qquad & \cot\left( \phi \right) & = & \dfrac{\text{adjacent}}{\text{opposite}} & = & \dfrac{2}{\sqrt{5}} \\[6pt] \end{array} \nonumber$$

 The result of the computation to find the missing side in Example $\PageIndex{ 1 }$ part (a) reveals one of the Pythagorean triples, $(5,12,13)$.

Definition: Pythagorean Triple

A Pythagorean triple consists of three positive integers $a$, $b$, and $c$, such that $a^2 + b^2 = c^2$.

There are an infinite number of Pythagorean triples, but the common ones found in Trigonometry textbooks are $(3,4,5)$, $(5, 12, 13)$, and $(7, 24, 25)$. Triangles having sides lengths matching Pythagorean triples get a special name.

Definition: Pythagorean Triangle

A right triangle whose sides form a Pythagorean triple is called a Pythagorean triangle.

It can be proved that every Pythagorean triangle must be a right triangle. For more information, see this Wikipedia page.

Checkpoint $\PageIndex{ 1 }$

Find the sine of the labeled angle in the triangle below. Round the answer to 4 decimal places.

Figure $\PageIndex{ 3 }$

Answer

$\sin\left(\alpha\right) \approx 0.7442$

The following example illustrates a necessary skill for us as we move forward in Trigonometry.

Example $\PageIndex{2}$

Write algebraic expressions for the six trigonometric ratios of the angle $\theta$ if the hypotenuse has length 2 and the side adjacent to $\theta$ has length $D$.

Solution

We begin by finding the missing side, which we shall call $x$. $$\begin{array}{rrclcr} & D^2 + x^2 & = & 2^2 & \quad & (\text{Pythagorean Theorem}) \\[6pt] \implies & D^2 + x^2 & = & 4 & \quad & \\[6pt] \implies & x^2 & = & 4 - D^2 & \quad & \\[6pt] \implies & x & = & \sqrt{4 - D^2} & \quad & \\[6pt] \end{array} \nonumber$$ We can now fill in the values of the trigonometric ratios. $$\begin{array}{rcccccrcccc} \sin\left(\theta\right) & = & \dfrac{\text { opposite }}{\text { hypotenuse }} & = & \dfrac{\sqrt{4-D^2}}{2} & \qquad & \csc \left(\theta\right) & = & \dfrac{\text { hypotenuse }}{\text { opposite }} & = & \dfrac{2}{\sqrt{4 - D^2}} \\[6pt] \cos\left(\theta\right) & = & \dfrac{\text { adjacent }}{\text { hypotenuse }} & = & \dfrac{D}{2} & \qquad & \sec \left(\theta\right) & = & \dfrac{\text { hypotenuse }}{\text { adjacent }} & = & \dfrac{2}{D} \\[6pt] \tan\left(\theta\right) & = & \dfrac{\text { opposite }}{\text { adjacent }} & = & \dfrac{\sqrt{4 - D^2}}{D} & \qquad & \cot \left(\theta\right) & = & \dfrac{\text { adjacent }}{\text { opposite }} & = & \dfrac{D}{\sqrt{4 - D^2}} \\[6pt] \end{array} \nonumber$$

Checkpoint $\PageIndex{2}$

If the hypotenuse of a right triangle is $h$ and the side opposite $\theta$ is $L$, write expressions in $h$ and $L$ for all six trigonometric functions of $\theta$.

Answer

$$\begin{array}{rcccrcc} \sin\left(\theta\right) & = & \dfrac{L}{h} & \qquad & \csc \left(\theta\right) & = & \dfrac{h}{L} \\[6pt] \cos\left(\theta\right) & = & \dfrac{\sqrt{h^2 - L^2}}{h} & \qquad & \sec \left(\theta\right) & = & \dfrac{h}{\sqrt{h^2 - L^2}} \\[6pt] \tan\left(\theta\right) & = & \dfrac{L}{\sqrt{h^2 - L^2}} & \qquad & \cot \left(\theta\right) & = & \dfrac{\sqrt{h^2 - L^2}}{h} \\[6pt] \end{array} \nonumber$$

Caution: Right Triangle Trigonometry Works For... Right Triangles

To calculate the trigonometric function of an angle within a triangle, it is crucial that the triangle be a right triangle. For example, in the triangle below, $\sin \left(\alpha\right)= \frac{4}{7}$, because $\triangle A B C$ is a right triangle. It is not true that $\sin \left(\alpha\right)=\frac{5}{7}$, or $\sin\left( \alpha\right)=\frac{6}{7}$.

Figure $\PageIndex{ 4 }$

Revisiting the Fundamental Identities

Our previous identities (the Ratio, Reciprocal, and Pythagorean Identities) still hold when using the right triangle definition of the trigonometric functions. For example, notice that $$\sin\left( \theta \right) = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{1}{\frac{\text{hypotenuse}}{\text{opposite}}} = \dfrac{1}{\csc\left( \theta \right)}. \nonumber$$ We leave it to the reader to show that all other Reciprocal Identities also hold.

Likewise, $$\tan\left( \theta \right) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{\frac{\text{opposite}}{\text{hypotenuse}}}{\frac{\text{adjacent}}{\text{hypotenuse}}} = \dfrac{\sin\left( \theta \right)}{\cos\left( \theta \right)}. \nonumber$$ Again, we leave it to the reader to finish showing that both Ratio Identities still hold.

Finally, it is left as a homework exercise to show that the Pythagorean Identity still holds with this new definition.

Example $\PageIndex{3}$

If $\theta$ is an acute angle in a right triangle and $\tan\left( \theta \right) = \frac{15}{8}$, compute the remaining trigonometric ratios of $\theta$.

Solution

It is best to sketch the right triangle with $\theta$ and where the side adjacent to $\theta$ has length 8 and the side opposite $\theta$ has length 15.

Figure $\PageIndex{ 5 }$

Using the Pythagorean Theorem, we get $$\begin{array}{rrcl} & 8^2 + 15^2 & = & h^2 \\[6pt] \implies & 289 & = & h^2 \\[6pt] \implies & 17 & = & h \\[6pt] \end{array} \nonumber$$ Hence, $$\begin{array}{rclcrcccccl} \sin\left( \theta \right) & = & \dfrac{15}{17} & \implies & \csc\left( \theta \right) & = & \dfrac{1}{\sin\left( \theta \right)} & = & \dfrac{1}{15/17} & = & \dfrac{17}{15} \\[6pt] \cos\left( \theta \right) & = & \dfrac{8}{17} & \implies & \sec\left( \theta \right) & = & \dfrac{1}{\cos\left( \theta \right)} & = & \dfrac{1}{8/17} & = & \dfrac{17}{8} \\[6pt] \tan\left( \theta \right) & = & \dfrac{15}{8} & \implies & \cot\left( \theta \right) & = & \dfrac{1}{\tan\left( \theta \right)} & = & \dfrac{1}{15/8} & = & \dfrac{8}{15} \\[6pt] \end{array} \nonumber$$

Checkpoint $\PageIndex{3}$

If $\beta$ is an acute angle in a right triangle and $\csc\left( \beta \right) = \frac{2}{\sqrt{3}}$, compute the remaining trigonometric ratios of $\beta$.

Answer

$$\begin{array}{rclcrcl} \sin\left( \beta \right) & = & \dfrac{\sqrt{3}}{2} & \qquad & \csc\left( \beta \right) & = & \dfrac{2}{\sqrt{3}} \\[6pt] \cos\left( \beta \right) & = & \dfrac{1}{2} & \qquad & \sec\left( \beta \right) & = & 2 \\[6pt] \tan\left( \beta \right) & = & \sqrt{3} & \quad & \cot\left( \beta \right) & = & \dfrac{1}{\sqrt{3}} \\[6pt] \end{array} \nonumber$$

Example $\PageIndex{ 4}$

If $\sin \left(\theta\right) = \frac{3}{7}$ and $\cos\left( \theta \right) = \frac{2\sqrt{10}}{7}$, where $\theta$ is an acute angle within a right triangle, find the value of $\cot \left(\theta\right)$ without drawing a triangle.

Solution

$$\begin{array}{rcl} \cot\left( \theta \right) & = & \dfrac{\cos\left( \theta \right)}{\sin\left( \theta \right)} \\[6pt] & = & \dfrac{2\sqrt{10}/7}{3/7} \\[6pt] & = & \dfrac{2\sqrt{10}}{3} \\[6pt] \end{array} \nonumber$$

Checkpoint $\PageIndex{ 4}$

If $\lambda$ is an acute angle in a right triangle, $\tan \left(\lambda \right) = \frac{2}{3}$, and $\cos \left( \lambda \right) = \frac{3}{\sqrt{13}}$, find $\sin\left( \lambda \right)$ using only the Ratio Identities.

Answer

$\sin\left( \lambda \right) = \frac{2}{\sqrt{13}}$

Right Triangle Trigonometry and the Special Angles

Now that we have a right triangle definition for the trigonometric functions, we can use our special triangles to help us quickly evaluate the trigonometric functions at $30^{ \circ }$, $45^{ \circ }$, and $60^{ \circ }$ angles. These angles are foundations upon which we build the rest of Trigonometry.

Table $\PageIndex{ 1 }$: Fundamental Trigonometric Ratios for the Special Angles

$$\begin{array}{|c|c|c|c|} \hline \text { Angle } & \text { Sine } & \text { Cosine } & \text { Tangent } \\[6pt] \hline 30^{\circ} & \frac{1}{2} & \frac{\sqrt{3}}{2} & \frac{1}{\sqrt{3}} \\[6pt] \hline 45^{\circ} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1 \\[6pt] \hline 60^{\circ} & \frac{\sqrt{3}}{2} & \frac{1}{2} & \sqrt{3} \\[6pt] \hline \end{array} \nonumber$$

The exact values for these trigonometric ratios should be committed to memory. A great way to remember them is to know the two special triangles in Figure $\PageIndex{ 6 }$ below. Writing down the three trigonometric ratios for the special angles from these triangles is simple.

Figure $\PageIndex{ 6 }$: Left: The $45^{\circ}-45^{\circ}-90^{\circ}$ special triangle, Right: The $30^{\circ}-60^{\circ}-90^{\circ}$ special triangle

For example, we can quickly compute $\tan\left( 60^{ \circ } \right)$ and $\sec\left( 30^{ \circ } \right)$ by sketching the $30^{ \circ }$-$60^{ \circ }$-$90^{ \circ }$ triangle and finding the ratios of the relevant sides.

$$\sec \left(30^{\circ}\right) = \frac{\text { hypotenuse }}{\text { adjacent }}=\frac{2}{\sqrt{3}} \quad \quad \quad \tan \left(60^{\circ}\right)=\frac{\text { opposite }}{\text { adjacent }}=\frac{\sqrt{3}}{1}=\sqrt{3} \nonumber$$

These are exact values for the trigonometric ratios, but we can also find decimal approximations. Use a calculator to verify the following approximate values.

$$\begin{array}{rcccc} & & \text{exact value} & & \text{approximation}\\[6pt] \sec \left(30^{\circ}\right) & = & \dfrac{2}{\sqrt{3}} & \approx & 1.1547 \\[6pt] \tan \left(60^{\circ}\right) & = & \sqrt{3} & \approx & 1.7321 \\[6pt] \end{array} \nonumber$$

Caution: Approximations are NOT Exact

It is essential to understand the difference between exact and approximate values. The decimal approximations above are rounded off. Even if a calculator shows ten or twelve digits, the values are not precisely correct - although they are adequate for most practical calculations.

Computational Note: To Rationalize or Not To Rationalize... That is the Question

Did you notice that we left $\sec\left(30^{\circ}\right) = \frac{2}{\sqrt{3}}$ with a radical in its denominator? This is quite common in Trigonometry. While one can rationalize the denominator to get $$\sec\left( 30^{\circ} \right) = \dfrac{2}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}, \nonumber$$ the need to do so with numerical expressions (expressions not involving variables) is an antiquated "throwback" to the days when calculators did not exist. As such, unless it truly simplifies an expression, we will be okay with leaving radicals in the denominators of numerical expressions for the rest of mathematics.

Example $\PageIndex{ 5 }$

Matterhorn Chocolate bars are sold in boxes shaped like triangular prisms. The two triangular ends of a box are equilateral triangles with 8-centimeter sides. When stacking the boxes on a shelf, the second level will be a triangle’s altitude above the bottom level. Find the exact length of the triangle’s altitude, $h$.

Solution

We have seen this question before (near the beginning of the textbook). The altitude divides the triangle into two $30^{\circ}-60^{\circ}-90^{\circ}$ right triangles, as shown in Figure $\PageIndex{ 7 }$ below.

Figure $\PageIndex{ 7 }$

The altitude is adjacent to the $30^{\circ}$ angle and the hypotenuse of the right triangle is 8 centimeters. Thus, using the fact that we already know $\cos\left( 30^{\circ} \right) = \frac{\sqrt{3}}{2}$, we get $$\begin{array}{rrclr} & \cos \left(30^{\circ}\right) & = & \dfrac{\text { adjacent }}{\text { hypotenuse }} & (\text{Fill in the values.}) \\[6pt] \implies & \dfrac{\sqrt{3}}{2} & = & \dfrac{h}{8} & \\[6pt] \implies & h & = & 8\left(\dfrac{\sqrt{3}}{2}\right)=4 \sqrt{3} & (\text{multiplying both sides by 8}) \\[6pt] \end{array} \nonumber$$ The altitude is exactly $4 \sqrt{3}$ centimeters long. From this exact answer, we can find approximations to any degree of accuracy we like. You can check that $4 \sqrt{3} \approx 6.9282$, so the altitude is approximately $6.9$ centimeters long.

Checkpoint $\PageIndex{ 5 }$

Show that $\cos^2\left( 60^{ \circ } \right) + \sin^2\left(60^{\circ}\right) = 1$.

Answer

$\cos^2\left( 60^{ \circ } \right) + \sin^2\left(60^{\circ}\right) = \left( \dfrac{1}{2} \right)^2 + \left( \dfrac{\sqrt{3}}{2} \right)^2 = \dfrac{1}{4} + \dfrac{3}{4} = 1$.

Example $\PageIndex{6}$

Let $x = 60^{ \circ }$ and $y=45^{ \circ }$. Evaluate. $$\sqrt{3} \csc\left( x \right) - \dfrac{5}{\sqrt{2}} \sec\left( y \right) + \sqrt{6} \tan\left( 90^{\circ} - x \right) \sin\left( y \right) \nonumber$$

Solution

Using our two special triangles in Figure $\PageIndex{ 6 }$, we get $$\begin{array}{rcl} \sqrt{3} \csc\left( x \right) - \dfrac{5}{\sqrt{2}} \sec\left( y \right) + \sqrt{6} \tan\left( 90^{\circ} - x \right) \sin\left( y \right) & = & \sqrt{3} \csc\left( 60^{ \circ }\right) - \dfrac{5}{\sqrt{2}} \sec\left( 45^{ \circ } \right) + \sqrt{6} \tan\left( 90^{\circ} - 60^{ \circ } \right) \sin\left( 45^{ \circ } \right) \\[6pt] & = & \sqrt{3} \csc\left( 60^{ \circ }\right) - \dfrac{5}{\sqrt{2}} \sec\left( 45^{ \circ } \right) + \sqrt{6} \tan\left( 30^{\circ} \right) \sin\left( 45^{ \circ } \right) \\[6pt] & = & \cancel{\sqrt{3}} \left( \dfrac{2}{\cancel{\sqrt{3}}}\right) - \left(\dfrac{5}{\cancel{\sqrt{2}}}\right) \left(\dfrac{\cancel{\sqrt{2}}}{1} \right) + \left(\cancel{\sqrt{6}}\right)\left(\dfrac{1}{\cancel{\sqrt{3}}}\right)\left(\dfrac{1}{\cancel{\sqrt{2}}}\right) \\[6pt] & = & 2 - 5 + 1 \\[6pt] & = & -2 \end{array} \nonumber$$

Checkpoint $\PageIndex{6}$

Evaluate and write as a single expression. $$3 \cos\left( 45^{ \circ } \right) - 2\cot\left( 30^{ \circ } \right) \nonumber$$

Answer

$\dfrac{3- 2 \sqrt{6}}{\sqrt{2}}$

Cofunction Identities

The following definition may answer a burning question you have had since being introduced to trigonometric functions.

Definition: Cofunction

The sine and cosine are called cofunctions. Likewise, the tangent and cotangent, as well as the secant and cosecant, are called cofunctions.

We say the sine is the cofunction of the cosine, and the cosine is the cofunction of the sine. The same holds for tangent and cotangent, and secant and cosecant. Now that we have a definition for cofunctions, what's the big deal?

The right triangle definition of the trigonometric functions reveals a new identity that was not so obvious when working with the coordinate definition of trigonometric functions. Consider the right triangle in Figure $\PageIndex{ 8 }$ below.

Figure $\PageIndex{ 8 }$

By the Triangle Sum, $\angle A + \angle B + \angle C = 180^{ \circ }$; however, $\angle C = 90^{ \circ }$. Therefore, $\angle A + \angle B = 90^{ \circ }$. That is, $\angle A = 90^{ \circ } - \angle B$. By the right triangle definition, $$\begin{array}{rcccccccl} \sin\left( \angle A \right) & = & \dfrac{\text{Opposite }A}{\text{Hypotenuse}} & = & \dfrac{\text{Adjacent }B}{\text{Hypotenuse}} & = & \cos\left( \angle B \right) & = & \cos\left( 90^{ \circ } - \angle A \right) \\[6pt] \sec\left( \angle A \right) & = & \dfrac{\text{Hypotenuse}}{\text{Adjacent }A} & = & \dfrac{\text{Hypotenuse}}{\text{Opposite }B} & = & \csc\left( \angle B \right) & = & \csc\left( 90^{ \circ } - \angle A \right) \\[6pt] \tan\left( \angle A \right) & = & \dfrac{\text{Opposite }A}{\text{Adjacent }A} & = & \dfrac{\text{Adjacent }B}{\text{Opposite }B} & = & \cot\left( \angle B \right) & = & \cot\left( 90^{ \circ } - \angle A \right) \\[6pt] \end{array} \nonumber$$

At this point, I hope you can see a budding relationship between a function and its cofunction. The following theorem summarizes the results (the proof of which will be held off until we have more mathematics to work with).

Theorem: Cofunction Identities

The trigonometric function of an angle is equal to the cofunction evaluated at the complement of the angle. That is, $$\begin{array}{rcccccccl} \sin\left( \theta \right) & = & \cos\left( 90^{ \circ } - \theta \right) & \quad & \text{and} & \quad & \cos\left( \theta \right) & = & \sin\left( 90^{ \circ } - \theta \right) \\[6pt] \sec\left( \theta \right) & = & \csc\left( 90^{ \circ } - \theta \right) & \quad & \text{and} & \quad & \csc\left( \theta \right) & = & \sec\left( 90^{ \circ } - \theta \right) \\[6pt] \tan\left( \theta \right) & = & \cot\left( 90^{ \circ } - \theta \right) & \quad & \text{and} & \quad & \cot\left( \theta \right) & = & \tan\left( 90^{ \circ } - \theta \right) \\[6pt] \end{array} \nonumber$$

Example $\PageIndex{7}$

Evaluate (by hand). $$\sin^2\left( 13^{ \circ } \right) + \sin^2\left( 77^{\circ} \right) \nonumber$$

Solution

$$\begin{array}{rcl} \sin^2\left( 13^{ \circ } \right) + \sin^2\left( 77^{ \circ } \right) & = & \sin^2\left( 13^{ \circ } \right) + \left(\sin\left( 77^{ \circ } \right)\right)^2 \\[6pt] & = & \sin^2\left( 13^{ \circ } \right) + \left(\cos\left(90^{ \circ } - 77^{ \circ } \right)\right)^2 \\[6pt] & = & \sin^2\left( 13^{ \circ } \right) + \left(\cos\left(13^{ \circ }\right)\right)^2 \\[6pt] & = & \sin^2\left( 13^{ \circ } \right) + \cos^2\left(13^{ \circ }\right) \\[6pt] & = & 1 \end{array} \nonumber$$

Checkpoint $\PageIndex{7}$

Write the function in terms of its cofunction.

1. $\sin\left( 13^{ \circ } \right)$
2. $\cot\left( 60^{ \circ } \right)$
3. $\sec\left( y \right)$

Answers

1. $\sin\left( 13^{ \circ } \right) = \cos\left( 90^{ \circ } - 13^{ \circ }\right) = \cos\left( 77^{ \circ } \right)$
2. $\cot\left( 60^{ \circ } \right) = \tan\left( 90^{ \circ } - 60^{ \circ } \right) = \tan\left( 30^{ \circ } \right)$
3. $\sec\left( y \right) = \csc\left( 90^{ \circ } - y \right)$

Observations of Values for Trigonometric Functions Using Right Triangle Trigonometry

We have now spent a decent amount of time with the trigonometric functions - both defined in terms of coordinates on the Cartesian plane and in terms of ratios of sides of right triangles. It's now time to formalize some observations.

Trigonometric Functions are Invariant When Scaling Right Triangles

The trigonometric function of an angle is the same for any right triangle, no matter the size or orientation of the triangle. Another way to say this is the values of the trigonometric functions are invariant under scaling of the right triangle. For example, Figure $\PageIndex{ 9 }$ (below) shows three right triangles, each with a $50^{\circ}$ angle. Although the sides of the triangle may be bigger or smaller, the ratio $\frac{\text { opposite }}{\text { hypotenuse }}$ is always the same for that angle because the triangles are similar. Therefore, $\sin\left( 50^{ \circ } \right)$ is the same for all three right triangles.

Figure $\PageIndex{ 9 }$

Similarly, the values of all the other trigonometric functions at $50^{ \circ }$ remain constant when we scale the triangles to be larger or smaller.

Some Trigonometric Functions Have Limited Ranges

Consider the right triangle definition of the trigonometric functions. Since the hypotenuse is the longest side of a right triangle, the ratios $$\sin\left( \theta \right) = \dfrac{\text{opposite}}{\text{hypotenuse}} \quad \text{and} \quad \cos\left( \theta \right) = \dfrac{\text{adjacent}}{\text{hypotenuse}} \nonumber$$ will always be less than 1.

Moreover, if we think about the coordinate definition of the trigonometric functions, $$\sin\left( \theta \right) = \dfrac{y}{r} \quad \text{and} \quad \cos\left( \theta \right) = \dfrac{x}{r}. \nonumber$$ Since $r = \sqrt{x^2 + y^2} \geq \sqrt{y^2} = |y|$ and $r = \sqrt{x^2 + y^2} \geq \sqrt{x^2} = |x|$, $$\begin{array}{rcccccccl} |y| \leq r & \implies & \dfrac{|y|}{r} \leq 1 & \implies & \left| \dfrac{y}{r} \right| \leq 1 & \implies & \left| \sin\left( \theta \right) \right| \leq 1 & \implies & -1 \leq \sin\left( \theta \right) \leq 1 \\[6pt] & & & & \text{and} & & & & \\[6pt] |x| \leq r & \implies & \dfrac{|x|}{r} \leq 1 & \implies & \left| \dfrac{x}{r} \right| \leq 1 & \implies & \left| \cos\left( \theta \right) \right| \leq 1 & \implies & -1 \leq \cos\left( \theta \right) \leq 1 \\[6pt] \end{array} \nonumber$$ The only times the sine or cosine become $1$ or $-1$ is when the given angle is a quadrantal angle.

Example $\PageIndex{8}$

If $\theta$ is an angle within a right triangle, is it possible for $\sec\left( \theta \right) = \frac{1}{2}$? Explain.

Solution

$\sec\left( \theta \right)$ is the ratio of the hypotenuse to the adjacent side of $\theta$; however, the hypotenuse of a right triangle is always longer than the other sides. Therefore, the ratio $\frac{\text{hypotenuse}}{\text{adjacent}}$ must be greater than 1 (by the discussion before this example, the secant can attain a value of 1 only if the given angle is a quadrantal angle).

Checkpoint $\PageIndex{8}$

Can $\csc\left( \theta \right)$ ever be 2?

Answer

Yes!

Using Special Angles to Approximate Trigonometric Functions at Other Angles

In a right triangle, as the angle $\theta$ increases, $\sin\left( \theta\right)$ increases but $\cos\left( \theta\right)$ decreases. Figure $\PageIndex{ 10 }$ (below) demonstrates why this is true. In each right triangle, the hypotenuse has the same length. However, as the angle increases, the opposite side gets longer, and the adjacent side gets shorter.

Figure $\PageIndex{ 10 }$

We can use the special angles as benchmarks for estimating and mental calculation. For example, we know that $\sin \left(60^{\circ}\right) \approx 0.8660$, so if $\sin \left(\theta\right) = 0.95$ for some unknown angle $\theta$ within a right triangle, we know that $\theta > 60^{\circ}$, because as $\theta$ increases, the sine of $\theta$ increases as well.

Example $\PageIndex{ 9 }$

If $\cos \left(\alpha\right) > \frac{\sqrt{3}}{2}$, where $\alpha$ is an angle within a right triangle, what can we say about $\alpha$?

Solution

As an angle increases from $0^{\circ}$ to $90^{\circ}$, its cosine decreases (why?). Now, $\cos \left(30^{\circ}\right) = \dfrac{\sqrt{3}}{2}$, so if $\cos \left(\alpha\right) > \frac{\sqrt{3}}{2}$, then $\alpha$ must be less than $30^{\circ}$.

Checkpoint $\PageIndex{ 9 }$

If $1 < \tan \left(\beta\right) < \sqrt{3}$, where $\beta$ is one of the angles in a right triangle, what can we say about $\beta$?

Answer

$45^{\circ} < \beta < 60^{\circ}$