# 9.1 Solving trigonometric equations with identities  (Page 3/9)

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Reciprocal Identities
$\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{csc}\text{\hspace{0.17em}}\theta }$ $\mathrm{csc}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}\theta }$
$\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{sec}\text{\hspace{0.17em}}\theta }$ $\mathrm{sec}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{cos}\text{\hspace{0.17em}}\theta }$
$\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{cot}\text{\hspace{0.17em}}\theta }$ $\mathrm{cot}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{tan}\text{\hspace{0.17em}}\theta }$

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
$\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }$ $\mathrm{cot}\text{\hspace{0.17em}}\theta =\frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{sin}\text{\hspace{0.17em}}\theta }$

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

## Summarizing trigonometric identities

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

${\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1$
$1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta$
$1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta$

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.

$\mathrm{tan}\left(-\theta \right)=-\mathrm{tan}\text{\hspace{0.17em}}\theta$
$\mathrm{cot}\left(-\theta \right)=-\mathrm{cot}\text{\hspace{0.17em}}\theta$
$\mathrm{sin}\left(-\theta \right)=-\mathrm{sin}\text{\hspace{0.17em}}\theta$
$\mathrm{csc}\left(-\theta \right)=-\mathrm{csc}\text{\hspace{0.17em}}\theta$
$\mathrm{cos}\left(-\theta \right)=\mathrm{cos}\text{\hspace{0.17em}}\theta$
$\mathrm{sec}\left(-\theta \right)=\mathrm{sec}\text{\hspace{0.17em}}\theta$

The reciprocal identities    define reciprocals of the trigonometric functions.

$\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{csc}\text{\hspace{0.17em}}\theta }$
$\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{sec}\text{\hspace{0.17em}}\theta }$
$\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{cot}\text{\hspace{0.17em}}\theta }$
$\mathrm{csc}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}\theta }$
$\mathrm{sec}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{cos}\text{\hspace{0.17em}}\theta }$
$\mathrm{cot}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{tan}\text{\hspace{0.17em}}\theta }$

The quotient identities    define the relationship among the trigonometric functions.

$\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }$
$\mathrm{cot}\text{\hspace{0.17em}}\theta =\frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{sin}\text{\hspace{0.17em}}\theta }$

## Graphing the equations of an identity

Graph both sides of the identity $\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}\theta =\frac{1}{\mathrm{tan}\text{\hspace{0.17em}}\theta }.\text{\hspace{0.17em}}$ In other words, on the graphing calculator, graph $\text{\hspace{0.17em}}y=\mathrm{cot}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=\frac{1}{\mathrm{tan}\text{\hspace{0.17em}}\theta }.$

Given a trigonometric identity, verify that it is true.

1. 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.
2. Look for opportunities to factor expressions, square a binomial, or add fractions.
3. Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.
4. If these steps do not yield the desired result, try converting all terms to sines and cosines.

## Verifying a trigonometric identity

Verify $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta \mathrm{cos}\text{\hspace{0.17em}}\theta =\mathrm{sin}\text{\hspace{0.17em}}\theta .$

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

$\begin{array}{ccc}\hfill \mathrm{tan}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta & =& \left(\frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }\right)\mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \\ & =& \left(\frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{\overline{)\mathrm{cos}\text{\hspace{0.17em}}\theta }}\right)\overline{)\mathrm{cos}\text{\hspace{0.17em}}\theta }\hfill \\ & =& \mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \end{array}$

Verify the identity $\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =1.$

$\begin{array}{ccc}\hfill \mathrm{csc}\text{\hspace{0.17em}}\theta \mathrm{cos}\text{\hspace{0.17em}}\theta \mathrm{tan}\text{\hspace{0.17em}}\theta & =& \left(\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}\theta }\right)\mathrm{cos}\text{\hspace{0.17em}}\theta \left(\frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }\right)\hfill \\ & =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{sin}\text{\hspace{0.17em}}\theta }\left(\frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }\right)\hfill \\ & =& \frac{\mathrm{sin}\text{\hspace{0.17em}}\theta \mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{sin}\text{\hspace{0.17em}}\theta \mathrm{cos}\text{\hspace{0.17em}}\theta }\hfill \\ & =& 1\hfill \end{array}$

## Verifying the equivalency using the even-odd identities

Verify the following equivalency using the even-odd identities:

$\left(1+\mathrm{sin}\text{\hspace{0.17em}}x\right)\left[1+\mathrm{sin}\left(-x\right)\right]={\mathrm{cos}}^{2}x$

Working on the left side of the equation, we have

## Verifying a trigonometric identity involving sec 2 θ

Verify the identity $\text{\hspace{0.17em}}\frac{{\mathrm{sec}}^{2}\theta -1}{{\mathrm{sec}}^{2}\theta }={\mathrm{sin}}^{2}\theta$

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

$\begin{array}{cccc}\hfill \frac{{\mathrm{sec}}^{2}\theta -1}{{\mathrm{sec}}^{2}\theta }& =& \frac{\left({\mathrm{tan}}^{2}\theta +1\right)-1}{{\mathrm{sec}}^{2}\theta }\hfill & {\phantom{\rule{2em}{0ex}}\text{sec}}^{2}\theta ={\mathrm{tan}}^{2}\theta +1\hfill \\ & =& \frac{{\mathrm{tan}}^{2}\theta }{{\mathrm{sec}}^{2}\theta }\hfill & \\ & =& {\mathrm{tan}}^{2}\theta \left(\frac{1}{{\mathrm{sec}}^{2}\theta }\right)& \\ & =& {\mathrm{tan}}^{2}\theta \left({\mathrm{cos}}^{2}\theta \right)\hfill & \phantom{\rule{2em}{0ex}}{\mathrm{cos}}^{2}\theta =\frac{1}{{\mathrm{sec}}^{2}\theta }\hfill \\ & =& \left(\frac{{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }\right)\left({\mathrm{cos}}^{2}\theta \right)\hfill & {\phantom{\rule{2em}{0ex}}\text{tan}}^{2}\theta =\frac{{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }\hfill \\ & =& \left(\frac{{\mathrm{sin}}^{2}\theta }{\overline{){\mathrm{cos}}^{2}\theta }}\right)\left(\overline{){\mathrm{cos}}^{2}\theta }\right)\hfill & \\ & =& {\mathrm{sin}}^{2}\theta \hfill & \end{array}$

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

$\begin{array}{ccc}\hfill \frac{{\mathrm{sec}}^{2}\theta -1}{{\mathrm{sec}}^{2}\theta }& =& \frac{{\mathrm{sec}}^{2}\theta }{{\mathrm{sec}}^{2}\theta }-\frac{1}{{\mathrm{sec}}^{2}\theta }\hfill \\ & =& 1-{\mathrm{cos}}^{2}\theta \hfill \\ & =& {\mathrm{sin}}^{2}\theta \hfill \end{array}$

Show that $\text{\hspace{0.17em}}\frac{\mathrm{cot}\text{\hspace{0.17em}}\theta }{\mathrm{csc}\text{\hspace{0.17em}}\theta }=\mathrm{cos}\text{\hspace{0.17em}}\theta .$

$\begin{array}{ccc}\hfill \frac{\mathrm{cot}\text{\hspace{0.17em}}\theta }{\mathrm{csc}\text{\hspace{0.17em}}\theta }& =& \frac{\frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{sin}\text{\hspace{0.17em}}\theta }}{\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}\theta }}\hfill \\ & =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{sin}\text{\hspace{0.17em}}\theta }\cdot \frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{1}\hfill \\ & =& \mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \end{array}$

## Creating and verifying an identity

Create an identity for the expression $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ by rewriting strictly in terms of sine.

There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:

Thus,

$2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}\theta =\frac{2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta }{1-{\mathrm{sin}}^{2}\text{\hspace{0.17em}}\theta }$

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