Learning Objectives
In this section, you will:
- Recall the 3 Pythagorean Identities.
- Apply the 3 Pythagorean Identities to find exact vaues.
- Recall the double-angle identities for sine and cosine.
- Apply double-angle identities to find exact values.
Let’s Get Started…
Recall that the equation of the unit circle is [latex]{x}^{2}+{y}^{2}=1[/latex] and that, on the unit circle, we define the coordinates of a point [latex]P[/latex] with respect to the angle [latex]t[/latex] measured from the positive [latex]x-[/latex]axis to the point [latex]P[/latex] by
[latex]x = cos\left(t\right)[/latex] and [latex]y=sin\left(t\right)[/latex].
Substituting these definitions of [latex]x[/latex] and [latex]y[/latex] into the equation of the unit circle gives us the Pythagorean Identity.
The Pythagorean Identity
[latex]{\mathrm{cos}}^{2}\left(t\right)+{\mathrm{sin}}^{2}\left(t\right)=1[/latex]
Alternate Forms of the Pythagorean Identity
We can use this fundamental identity, plus the definitions of all 6 trigonometric functions, to derive alternate forms of the Pythagorean Identity, [latex]\,{\mathrm{cos}}^{2}\left(t\right)+{\mathrm{sin}}^{2}\left(t\right)=1.\,[/latex] One form is obtained by dividing both sides by [latex]\,{\mathrm{cos}}^{2}\left(t\right).[/latex]
[latex]\begin{array}{ccc}\hfill \frac{{\mathrm{cos}}^{2}\left(t\right)}{{\mathrm{cos}}^{2}\left(t\right)}+\frac{{\mathrm{sin}}^{2}\left(t\right)}{{\mathrm{cos}}^{2}\left(t\right)}& =& \frac{1}{{\mathrm{cos}}^{2}\left(t\right)}\hfill \\ \hfill & & & \hfill\\ \hfill 1+{\mathrm{tan}}^{2}\left(t\right)& =& {\mathrm{sec}}^{2}\left(t\right)\hfill \end{array}[/latex]
The other form is obtained by dividing both sides by [latex]\,{\mathrm{sin}}^{2}\left(t\right).[/latex]
[latex]\begin{array}{ccc}\hfill \frac{{\mathrm{cos}}^{2}\left(t\right)}{{\mathrm{sin}}^{2}\left(t\right)}+\frac{{\mathrm{sin}}^{2}\left(t\right)}{{\mathrm{sin}}^{2}\left(t\right)}& =& \frac{1}{{\mathrm{sin}}^{2}\left(t\right)}\hfill \\ \hfill & & & \hfill \\ \hfill {\mathrm{cot}}^{2}\left(t\right)+1& =& {\mathrm{csc}}^{2}\left(t\right)\hfill \end{array}[/latex]
Alternate Forms of the Pythagorean Identity
EXAMPLE 1
Using Identities to Relate Trigonometric Functions
If [latex]\,\mathrm{cos}\left(t\right)=\frac{12}{13}\,[/latex] and [latex]\,t\,[/latex] is in quadrant IV, as shown in Figure 1, find the values of the other five trigonometric functions.
Figure 1
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Solution
Use [latex]\,{\mathrm{cos}}^{2}\left(t\right)+{\mathrm{sin}}^{2}\left(t\right)=1[/latex] and the definitions of the remaining functions by relating them to sine and cosine.
The sign of the sine depends on the [latex]y-[/latex]values in the quadrant where the angle is located. Since the angle is in quadrant IV, where the [latex]y-[/latex]values are negative, its sine is negative, [latex]\,-\frac{5}{13}.[/latex]
The remaining functions can be calculated using identities relating them to sine and cosine.
[latex]\begin{array}{cccc}\hfill \text{tan }\left(t\right)& =\frac{\text{sin }\left(t\right)}{\text{cos }\left(t\right)}\hfill & =\frac{-\frac{5}{13}}{\frac{12}{13}}\hfill & =-\frac{5}{12}\hfill \\ \hfill \text{sec }\left(t\right)& =\frac{1}{\text{cos }\left(t\right)}\hfill & =\frac{1}{\frac{12}{13}}\hfill & =\frac{13}{12}\hfill \\ \hfill \text{csc }\left(t\right)& =\frac{1}{\text{sin }\left(t\right)}\hfill & =\frac{1}{-\frac{5}{13}}\hfill & =\frac{-13}{5}\hfill \\ \hfill \text{cot }\left(t\right)& =\frac{1}{\text{tan }\left(t\right)}\hfill & =\frac{1}{-\frac{5}{12}}\hfill & =-\frac{12}{5}\hfill \end{array}[/latex]
Try It #1
If [latex]\,\mathrm{sec}\left(t\right)=-\frac{17}{8}\,[/latex] and [latex]\,0 \lt t \lt \pi ,[/latex] find the values of the other five functions.
EXAMPLE 2
Finding the Values of Trigonometric Functions
Find the values of the six trigonometric functions of angle [latex]\,t\,[/latex] based on Figure 2.
Figure 2
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Solution
[latex]\begin{array}{cccc}\hfill \text{sin }\left(t\right)& =y\hfill & =-\frac{\sqrt{3}}{2}\hfill & \\ \hfill \text{cos }\left(t\right)& =x\hfill & =-\frac{1}{2}\hfill & \\ \hfill \text{tan }\left(t\right)& =\frac{\text{sin }\left(t\right)}{\text{cos }\left(t\right)}\hfill & =\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\hfill & =\sqrt{3}\hfill \\ \hfill \text{sec }\left(t\right)& =\frac{1}{\text{cos }\left(t\right)}\hfill & =\frac{1}{-\frac{1}{2}}\hfill & =-2\hfill \\ \hfill \text{csc }\left(t\right)& =\frac{1}{\text{sin }\left(t\right)}\hfill & =\frac{1}{-\frac{\sqrt{3}}{2}}\hfill & =-\frac{2\sqrt{3}}{3}\hfill \\ \hfill \text{cot }\left(t\right)& =\frac{1}{\text{tan }\left(t\right)}\hfill & =\frac{1}{\sqrt{3}}\hfill & =\frac{\sqrt{3}}{3}\hfill \end{array}[/latex]
Try It #2
Find the values of the six trigonometric functions of angle [latex]\,t\,[/latex] based on Figure 3.
Figure 3
EXAMPLE 3
Finding the Value of Trigonometric Functions
If [latex]\,\mathrm{sin}\left(t\right)=-\frac{\sqrt{3}}{2}\,\text{and}\,\text{cos}\left(t\right)=\frac{1}{2},\text{find}\,\text{sec}\left(t\right),\text{csc}\left(t\right),\text{tan}\left(t\right),\text{cot}\left(t\right).[/latex]
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Solution
[latex]\begin{array}{ccc}\hfill \text{sec }\left(t\right)& =\frac{1}{\text{cos }\left(t\right)}\hfill & =\frac{1}{\frac{1}{2}}=2\hfill \\ \hfill \text{csc }\left(t\right)& =\frac{1}{\text{sin }\left(t\right)}\hfill & =\frac{1}{-\frac{\sqrt{3}}{2}}-\frac{2\sqrt{3}}{3}\hfill \\ \hfill \text{tan }\left(t\right)& =\frac{\text{sin }\left(t\right)}{\text{cos }\left(t\right)}\hfill & =\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\hfill \\ \hfill \text{cot }\left(t\right)& =\frac{1}{\text{tan }\left(t\right)}\hfill & =\frac{1}{-\sqrt{3}}=-\frac{\sqrt{3}}{3}\hfill \end{array}[/latex]
Try It #3
[latex]\,\mathrm{sin}\left(t\right)=\frac{\sqrt{2}}{2}\,\text{and}\,\mathrm{cos}\left(t\right)=\frac{\sqrt{2}}{2},\text{find}\,\text{sec}\left(t\right),\text{csc}\left(t\right),\text{tan}\left(t\right),\text{and}\,\text{cot}\left(t\right)[/latex]
Double-Angle Identities
Figure 4
Bicycle ramps made for competition see Figure 4 must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be [latex]\theta[/latex] such that [latex]tan\left(\theta\right)=\frac{5}{3}[/latex]. The angle is divided in half for novices. What is the steepness of the ramp for novices? Now, we will investigate three additional categories of identities that we can use to answer questions such as this one.
Using Double-Angle Formulas to Find Exact Values
The double-angle formulas are a special case of the sum formulas for sine and cosine, where [latex]\alpha=\beta[/latex]. Deriving the double-angle formula for sine begins with its sum formula,
[latex]sin\left(\alpha+\beta\right)=sin\left(\alpha\right)cos\left(\beta\right)+cos\left(\alpha\right)sin\left(\beta\right)[/latex]
If we let [latex]\alpha=\beta=\theta[/latex], then we have
[latex]sin\left(\theta+\theta\right)=sin\left(\theta\right)cos\left(\theta\right)+cos\left(\theta\right)sin\left(\theta\right)[/latex]
or
[latex]sin\left(2\theta\right)=2sin\left(\theta\right)cos\left(\theta\right)[/latex]
Deriving the double-angle for cosine gives us three options. First, starting from the sum formula,
[latex]cos\left(\alpha+\beta\right)=cos\left(\alpha\right)cos\left(\beta\right)-sin\left(\alpha\right)sin\left(\beta\right)[/latex],
and letting [latex]\alpha=\beta=\theta[/latex], we have
[latex]cos\left(\theta+\theta\right)=cos\left(\theta\right)cos\left(\theta\right)-sin\left(\theta\right)sin\left(\theta\right)[/latex]
or
[latex]cos\left(2\theta\right)={cos}^{2}\left(\theta\right)-{sin}^{2}\left(\theta\right)[/latex]
Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:
[latex]cos\left(2\theta\right)={cos}^{2}\left(\theta\right)-{sin}^{2}\left(\theta\right)=\left(1-{sin}^{2}\left(\theta\right)\right)-{sin}^{2}\left(\theta\right)[/latex]
or
[latex]cos\left(2\theta\right)=1-2{sin}^{2}\left(\theta\right)[/latex]
The second interpretation is:
[latex]cos\left(2\theta\right)={cos}^{2}\left(\theta\right)-{sin}^{2}\left(\theta\right)={cos}^{2}\left(\theta\right)-\left(1-{cos}^{2}\left(\theta\right)\right)[/latex]
or
[latex]cos\left(2\theta\right)=2{cos}^{2}\left(\theta\right)-1[/latex]
Double-Angle Formulas
The double-angle formulas for sine and cosine are summarized as follows:
[latex]\begin{array}{ccc}\hfill sin\left(2\theta\right) & = & 2sin\left(\theta\right)cos\left(\theta\right) \hfill \\ \hfill & & & \hfill \\ \hfill \text{and} & & \hfill \\ \hfill & & & \hfill \\ \hfill cos\left(2\theta\right) & = & {cos}^{2}\left(\theta\right)-{sin}^{2}\left(\theta\right) \hfill \\ \hfill & = & 1-2{sin}^{2}\left(\theta\right)\hfill \\ \hfill & = & cos\left(2\theta\right)=2{cos}^{2}\left(\theta\right)-1\hfill \end{array}[/latex]
How To
Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.
- Draw a triangle to reflect the given information.
- Determine the correct double-angle formula.
- Substitute values into the formula based on the triangle.
- Simplify.
EXAMPLE 4
Using a Double-Angle Formula to Find the Exact Value Involving Tangent
Given that [latex]tan\left(\theta\right)=-\frac{3}{4}[/latex] and [latex]\theta[/latex] is in Quadrant II, find [latex]sin\left(2\theta\right)[/latex] and [latex]cos\left(2\theta\right)[/latex].
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Solution
If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given [latex]tan\left(\theta\right)=-\frac{3}{4}[/latex], such that [latex]\theta[/latex] is in Quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because [latex]\theta[/latex] is in the second quadrant, the adjacent side is on the [latex]x-[/latex]axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:
[latex]\begin{array}{ccc}\hfill {\left(-4\right)}^{2}+{\left(3\right)}^{2} & = & {c}^{2} \hfill \\ \hfill 16+9 & = & {c}^{2} \hfill \\ \hfill 25 & = & {c}^{2} \hfill \\ \hfill 5 & = & c \hfill \end{array}[/latex]
Now we can draw a triangle similar to the one shown in Figure 5.
Figure 5
Let’s begin by writing the double-angle formula for sine.
[latex]sin\left(2\theta\right)=2sin\left(\theta\right)cos\left(\theta\right)[/latex]
We see that we to need to find [latex]sin\left(\theta\right)[/latex] and [latex]cos\left(\theta\right)[/latex]. Based on Figure 5, we see that the hypotenuse equals 5, so [latex]sin\left(\theta\right)=\frac{3}{5}[/latex] and [latex]cos\left(\theta\right)=-\frac{4}{5}[/latex].
Now we can substitute these values into the double-angle identity for sine, and simplify.
[latex]sin\left(2\theta\right)=2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right)[/latex]
Thus,
[latex]sin\left(2\theta\right)=-\frac{24}{25}[/latex].
It’s arbitrary which double-angle identity we use for cosine. Let’s use the first one:
[latex]cos\left(2\theta\right)={cos}^{2}\left(\theta\right)-{sin}^{2}\left(\theta\right)[/latex].
Again, substitute the values of the sine and cosine into the equation, and simplify.
[latex]cos\left(2\theta\right)={\left(-\frac{4}{5}\right)}^{2}-{\left(\frac{3}{5}\right)}^{2}=\frac{16}{25}-\frac{9}{25}[/latex]
So we know
[latex]cos\left(2\theta\right)=\frac{7}{25}[/latex].
Try It #4
Given [latex]sin\left(\alpha\right)=\frac{5}{8}[/latex] and [latex]\alpha[/latex] is in Quadrant I, find [latex]cos\left(2\alpha\right)[/latex].
EXAMPLE 5
Using the Double-Angle Formula for Cosine without Exact Values
Use the double-angle formula for cosine to write [latex]cos\left(6x\right)[/latex] in terms of [latex]cos\left(3x\right)[/latex].
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Solution
Since everything in this problem is in terms of cosine, let’s use the third version of the double-angle identity for cosine:
[latex]cos\left(2\theta\right)=2{cos}^{2}\left(\theta\right)-1[/latex].
Here, we can write [latex]cos\left(6x\right)=cos\left(2\left(3x\right)\right)[/latex]. Then, let [latex]\theta=3x[/latex] in the identity above.
[latex]\begin{array}{ccc}\hfill cos\left(6x\right) & = & cos\left(2\left(3x\right)\right) \hfill \\ \hfill & = & 2{cos}^{2}\left(3x\right)-1 \hfill \end{array}[/latex]
Analysis
This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.
Section 2.4 Exercises
[Answers to odd problem numbers are provided at the end of the problem set. Just scroll down!]
These problems are from Stitz & Zeager’s open Precalculus textbook, version 3
1. [latex]sin\left(\theta\right)=\frac{3}{5}[/latex] with [latex]\theta[/latex] in Quadrant II
2. [latex]tan\left(\theta\right)=\frac{12}{5}[/latex] with [latex]\theta[/latex] in Quadrant III
3. [latex]csc\left(\theta\right)=\frac{25}{24}[/latex] with [latex]\theta[/latex] in Quadrant I
4. [latex]sec\left(\theta\right)=7[/latex] with [latex]\theta[/latex] in Quadrant IV
5. [latex]csc\left(\theta\right)=-\frac{10\sqrt{91}}{91}[/latex] with [latex]\theta[/latex] in Quadrant I
6. [latex]cot\left(\theta\right)=-23[/latex] with [latex]\theta[/latex] in Quadrant II
7. [latex]tan\left(\theta\right)=-2[/latex] with [latex]\theta[/latex] in Quadrant IV
8. [latex]sec\left(\theta\right)=-4[/latex] with [latex]\theta[/latex] in Quadrant II
9. [latex]cot\left(\theta\right)=\sqrt{5}[/latex] with [latex]\theta[/latex] in Quadrant III
10. [latex]csc\left(\theta\right)=\frac{1}{3}[/latex] with [latex]\theta[/latex] in Quadrant I
11. [latex]cot\left(\theta\right)=2[/latex] with [latex]0 \lt \theta \lt \frac{\pi}{2}.[/latex]
12. [latex]csc\left(\theta\right)=5[/latex] with [latex]\frac{\pi}{2} \lt \theta \lt \pi.[/latex]
13. [latex]tan\left(\theta\right)=\sqrt{10}[/latex] with [latex]\pi \lt \theta \lt \frac{3\pi}{2}.[/latex]
14. [latex]sec\left(\theta\right)=2\sqrt{5}[/latex] with [latex]\frac{3\pi}{2} \lt \theta \lt 2\pi.[/latex]
These problems are from OpenStax’sPrecalculus, 2nd edition textbook.
For the following exercises, find the exact values of [latex]sin\left(2\theta\right)[/latex] and [latex]cos\left(2\theta\right)[/latex] without solving for [latex]x[/latex]…
15. If [latex]sin\left(x\right)=\frac{1}{8}[/latex] and [latex]x[/latex] is in Quadrant I.
16. If [latex]cos\left(x\right)=\frac{2}{3}[/latex] and [latex]x[/latex] is in Quadrant I.
17. If [latex]cos\left(x\right)=-\frac{1}{2}[/latex] and [latex]x[/latex] is in Quadrant III.
18. If [latex]tan\left(x\right)=-8[/latex] and [latex]x[/latex] is in Quadrant IV.
For the following exercises, find the values of the six trigonometric functions if the conditions provided hold.
