7.1 Solving trigonometric equations with identities (Page 3/9)

Precalculus Page 3 / 9

Reciprocal Identities
$\sin θ = \frac{1}{\csc θ}$	$\csc θ = \frac{1}{\sin θ}$
$\cos θ = \frac{1}{\sec θ}$	$\sec θ = \frac{1}{\cos θ}$
$\tan θ = \frac{1}{\cot θ}$	$\cot θ = \frac{1}{\tan θ}$

The final set of identities is the set of quotient identities , which define relationships among certain trigonometric functions and can be very helpful in verifying other identities. See [link] .


Quotient Identities
$\tan θ = \frac{\sin θ}{\cos θ}$	$\cot θ = \frac{\cos θ}{\sin θ}$

The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.

A General Note

Summarizing trigonometric identities

The Pythagorean identities are based on the properties of a right triangle.

\cos^{2} θ + \sin^{2} θ = 1

1 + \cot^{2} θ = \csc^{2} θ

1 + \tan^{2} θ = \sec^{2} θ

The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.

\tan (- θ) = - \tan θ

\cot (- θ) = - \cot θ

\sin (- θ) = - \sin θ

\csc (- θ) = - \csc θ

\cos (- θ) = \cos θ

\sec (- θ) = \sec θ

The reciprocal identities define reciprocals of the trigonometric functions.

\sin θ = \frac{1}{\csc θ}

\cos θ = \frac{1}{\sec θ}

\tan θ = \frac{1}{\cot θ}

\csc θ = \frac{1}{\sin θ}

\sec θ = \frac{1}{\cos θ}

\cot θ = \frac{1}{\tan θ}

The quotient identities define the relationship among the trigonometric functions.

\tan θ = \frac{\sin θ}{\cos θ}

\cot θ = \frac{\cos θ}{\sin θ}

Graphing the equations of an identity

Graph both sides of the identity $\cot θ = \frac{1}{\tan θ} .$ In other words, on the graphing calculator, graph $y = \cot θ$ and $y = \frac{1}{\tan θ} .$

See [link] .

Graph of y = cot(theta) and y=1/tan(theta) from -2pi to 2pi. They are the same!

Got questions? Get instant answers now!

How To

Given a trigonometric identity, verify that it is true.

Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.
Look for opportunities to factor expressions, square a binomial, or add fractions.
Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.
If these steps do not yield the desired result, try converting all terms to sines and cosines.

Verifying a trigonometric identity

Verify $\tan θ \cos θ = \sin θ .$

We will start on the left side, as it is the more complicated side:

\begin{array}{l} \tan θ \cos θ = (\frac{\sin θ}{\cos θ}) \cos θ \\ = (\frac{\sin θ}{\cos θ}) \cos θ \\ = \sin θ \end{array}

Got questions? Get instant answers now!

Try It

Verify the identity $\csc θ \cos θ \tan θ = 1.$

\begin{array}{l} \csc θ \cos θ \tan θ = (\frac{1}{\sin θ}) \cos θ (\frac{\sin θ}{\cos θ}) \\ = \frac{\cos θ}{\sin θ} (\frac{\sin θ}{\cos θ}) \\ = \frac{\sin θ \cos θ}{\sin θ \cos θ} \\ = 1 \end{array}

Got questions? Get instant answers now!

Verifying the equivalency using the even-odd identities

Verify the following equivalency using the even-odd identities:

(1 + \sin x) [1 + \sin (- x)] = \cos^{2} x

Working on the left side of the equation, we have

\begin{array}{l} (1 + \sin x) [1 + \sin (- x)] = (1 + \sin x) (1 - \sin x) & Since sin(− x)= - \sin x \\ = 1 - \sin^{2} x & Difference of squares \\ = \cos^{2} x & {cos}^{2} x = 1 - \sin^{2} x \end{array}

Got questions? Get instant answers now!

Verifying a trigonometric identity involving sec ² θ

Verify the identity $\frac{\sec^{2} θ - 1}{\sec^{2} θ} = \sin^{2} θ$

As the left side is more complicated, let’s begin there.

\begin{array}{l} \frac{\sec^{2} θ - 1}{\sec^{2} θ} = \frac{(\tan^{2} θ + 1) - 1}{\sec^{2} θ} & {sec}^{2} θ = \tan^{2} θ + 1 \\ = \frac{\tan^{2} θ}{\sec^{2} θ} \\ = \tan^{2} θ (\frac{1}{\sec^{2} θ}) \\ = \tan^{2} θ (\cos^{2} θ) & \cos^{2} θ = \frac{1}{\sec^{2} θ} \\ = (\frac{\sin^{2} θ}{\cos^{2} θ}) (\cos^{2} θ) & \begin{matrix} \begin{matrix}  \end{matrix} \end{matrix} & {tan}^{2} θ = \frac{\sin^{2} θ}{\cos^{2} θ} \\ = (\frac{\sin^{2} θ}{\cos^{2} θ}) (\cos^{2} θ) \\ = \sin^{2} θ \end{array}

There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

\begin{array}{l} \frac{\sec^{2} θ - 1}{\sec^{2} θ} = \frac{\sec^{2} θ}{\sec^{2} θ} - \frac{1}{\sec^{2} θ} \\ = 1 - \cos^{2} θ \\ = \sin^{2} θ \end{array}

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Practice Key Terms 4

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Precalculus' conversation and receive update notifications?

Ask

	4 Solving trigonometric equations with identities By OpenStax Read Online Course
	2 Solving trigonometric equations with identities By OpenStax Read Online Course
	My OCA Mock By Mike Wolf Start Exam
	6 AP 06 Skeletal System Essay By OpenStax Start Flashcards
	9 BOD- Liver Quiz .... By Brooke Delaney Start Quiz
	Young Economist MCQ Test By Robert Murphy Start Test
©flickr: quinn	Neuroanatomy By George Turner Start Quiz
	2 Muscular System MCQ By Nick Swain Start Quiz
	2 AP 02 Chemical Level of Organization MCQ By OpenStax Start Quiz
©flickr: eLife	Mycobacterial Infections (Mandell 8th Ch 251-254) By Cath Yu Start Quiz
©flickr: Iqbal	Liver Cancer By Darlene Paliswat Start Test
	8 AP 08 Appendicular Skeleton Essay By OpenStax Start Flashcards

7.1 Solving trigonometric equations with identities (Page 3/9)

Summarizing trigonometric identities

Graphing the equations of an identity

Verifying a trigonometric identity

Verifying the equivalency using the even-odd identities

Verifying a trigonometric identity involving sec 2 θ

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Verifying a trigonometric identity involving sec ² θ