# 9.1 Solving trigonometric equations with identities  (Page 5/9)

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## Simplify by rewriting and using substitution

Simplify the expression by rewriting and using identities:

${\mathrm{csc}}^{2}\theta -{\mathrm{cot}}^{2}\theta$

We can start with the Pythagorean identity.

$1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta$

Now we can simplify by substituting $\text{\hspace{0.17em}}1+{\mathrm{cot}}^{2}\theta \text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}{\mathrm{csc}}^{2}\theta .\text{\hspace{0.17em}}$ We have

$\begin{array}{ccc}\hfill {\mathrm{csc}}^{2}\theta -{\mathrm{cot}}^{2}\theta & =& 1+{\mathrm{cot}}^{2}\theta -{\mathrm{cot}}^{2}\theta \hfill \\ & =& 1\hfill \end{array}$

Use algebraic techniques to verify the identity: $\text{\hspace{0.17em}}\frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{1+\mathrm{sin}\text{\hspace{0.17em}}\theta }=\frac{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }.$

(Hint: Multiply the numerator and denominator on the left side by $\text{\hspace{0.17em}}1-\mathrm{sin}\text{\hspace{0.17em}}\theta .\right)$

$\begin{array}{ccc}\hfill \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{1+\mathrm{sin}\text{\hspace{0.17em}}\theta }\left(\frac{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }\right)& =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta \left(1-\mathrm{sin}\text{\hspace{0.17em}}\theta \right)}{1-{\mathrm{sin}}^{2}\theta }\hfill \\ & =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta \left(1-\mathrm{sin}\text{\hspace{0.17em}}\theta \right)}{{\mathrm{cos}}^{2}\theta }\hfill \\ & =& \frac{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }\hfill \end{array}$

Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

## Key equations

 Pythagorean identities $\begin{array}{l}{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1\\ 1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta \\ 1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta \end{array}$ Even-odd identities $\begin{array}{ccc}\mathrm{tan}\left(-\theta \right)& =& -\mathrm{tan}\text{\hspace{0.17em}}\theta \\ \mathrm{cot}\left(-\theta \right)& =& -\mathrm{cot}\text{\hspace{0.17em}}\theta \\ \mathrm{sin}\left(-\theta \right)& =& -\mathrm{sin}\text{\hspace{0.17em}}\theta \\ \mathrm{csc}\left(-\theta \right)& =& -\mathrm{csc}\text{\hspace{0.17em}}\theta \\ \mathrm{cos}\left(-\theta \right)& =& \mathrm{cos}\text{\hspace{0.17em}}\theta \\ \mathrm{sec}\left(-\theta \right)& =& \mathrm{sec}\text{\hspace{0.17em}}\theta \end{array}$ Reciprocal identities $\begin{array}{ccc}\mathrm{sin}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{csc}\text{\hspace{0.17em}}\theta }\\ \mathrm{cos}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{sec}\text{\hspace{0.17em}}\theta }\\ \mathrm{tan}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{cot}\text{\hspace{0.17em}}\theta }\\ \mathrm{csc}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{sin}\text{\hspace{0.17em}}\theta }\\ \mathrm{sec}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{cos}\text{\hspace{0.17em}}\theta }\\ \mathrm{cot}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{tan}\text{\hspace{0.17em}}\theta }\end{array}$ Quotient identities $\begin{array}{ccc}\mathrm{tan}\text{\hspace{0.17em}}\theta & =& \frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }\\ \mathrm{cot}\text{\hspace{0.17em}}\theta & =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{sin}\text{\hspace{0.17em}}\theta }\end{array}$

## Key concepts

• There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.
• Graphing both sides of an identity will verify it. See [link] .
• Simplifying one side of the equation to equal the other side is another method for verifying an identity. See [link] and [link] .
• The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See [link] .
• We can create an identity and then verify it. See [link] .
• Verifying an identity may involve algebra with the fundamental identities. See [link] and [link] .
• Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics. See [link] , [link] , and [link] .

## Verbal

We know $\text{\hspace{0.17em}}g\left(x\right)=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is an even function, and $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\left(x\right)=\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ are odd functions. What about $\text{\hspace{0.17em}}G\left(x\right)={\mathrm{cos}}^{2}x,F\left(x\right)={\mathrm{sin}}^{2}x,$ and $\text{\hspace{0.17em}}H\left(x\right)={\mathrm{tan}}^{2}x?\text{\hspace{0.17em}}$ Are they even, odd, or neither? Why?

All three functions, $\text{\hspace{0.17em}}F,G,$ and $H,$ are even.

This is because $\text{\hspace{0.17em}}F\left(-x\right)=\mathrm{sin}\left(-x\right)\mathrm{sin}\left(-x\right)=\left(-\mathrm{sin}\text{\hspace{0.17em}}x\right)\left(-\mathrm{sin}\text{\hspace{0.17em}}x\right)={\mathrm{sin}}^{2}x=F\left(x\right),G\left(-x\right)=\mathrm{cos}\left(-x\right)\mathrm{cos}\left(-x\right)=\mathrm{cos}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x={\mathrm{cos}}^{2}x=G\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}H\left(-x\right)=\mathrm{tan}\left(-x\right)\mathrm{tan}\left(-x\right)=\left(-\mathrm{tan}\text{\hspace{0.17em}}x\right)\left(-\mathrm{tan}\text{\hspace{0.17em}}x\right)={\mathrm{tan}}^{2}x=H\left(x\right).$

Examine the graph of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ on the interval $\text{\hspace{0.17em}}\left[-\pi ,\pi \right].\text{\hspace{0.17em}}$ How can we tell whether the function is even or odd by only observing the graph of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sec}\text{\hspace{0.17em}}x?$

After examining the reciprocal identity for $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}t,$ explain why the function is undefined at certain points.

When $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t=0,$ then $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}t=\frac{1}{0},$ which is undefined.

All of the Pythagorean identities are related. Describe how to manipulate the equations to get from $\text{\hspace{0.17em}}{\mathrm{sin}}^{2}t+{\mathrm{cos}}^{2}t=1\text{\hspace{0.17em}}$ to the other forms.

## Algebraic

For the following exercises, use the fundamental identities to fully simplify the expression.

$\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{sin}\text{\hspace{0.17em}}x$

$\mathrm{sin}\left(-x\right)\text{\hspace{0.17em}}\mathrm{cos}\left(-x\right)\text{\hspace{0.17em}}\mathrm{csc}\left(-x\right)$

$\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}{\mathrm{cos}}^{2}x$

$\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{csc}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cot}\left(-x\right)$

$\frac{\mathrm{cot}\text{\hspace{0.17em}}t+\mathrm{tan}\text{\hspace{0.17em}}t}{\mathrm{sec}\left(-t\right)}$

$\mathrm{csc}\text{\hspace{0.17em}}t$

$3\text{\hspace{0.17em}}{\mathrm{sin}}^{3}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}t+{\mathrm{cos}}^{2}\text{\hspace{0.17em}}t+2\text{\hspace{0.17em}}\mathrm{cos}\left(-t\right)\mathrm{cos}\text{\hspace{0.17em}}t$

$-\mathrm{tan}\left(-x\right)\mathrm{cot}\left(-x\right)$

$-1$

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

$\frac{1+{\mathrm{tan}}^{2}\theta }{{\mathrm{csc}}^{2}\theta }+{\mathrm{sin}}^{2}\theta +\frac{1}{{\mathrm{sec}}^{2}\theta }$

${\mathrm{sec}}^{2}x$

$\left(\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{{\mathrm{csc}}^{2}x}+\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{{\mathrm{sec}}^{2}x}\right)\left(\frac{1+\mathrm{tan}\text{\hspace{0.17em}}x}{1+\mathrm{cot}\text{\hspace{0.17em}}x}\right)-\frac{1}{{\mathrm{cos}}^{2}x}$

$\frac{1-{\mathrm{cos}}^{2}\text{\hspace{0.17em}}x}{{\mathrm{tan}}^{2}\text{\hspace{0.17em}}x}+2\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}x$

${\mathrm{sin}}^{2}x+1$

For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression.

$\frac{\mathrm{tan}\text{\hspace{0.17em}}x+\mathrm{cot}\text{\hspace{0.17em}}x}{\mathrm{csc}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x$

$\frac{\mathrm{sec}\text{\hspace{0.17em}}x+\mathrm{csc}\text{\hspace{0.17em}}x}{1+\mathrm{tan}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x}$

$\frac{\mathrm{cos}\text{\hspace{0.17em}}x}{1+\mathrm{sin}\text{\hspace{0.17em}}x}+\mathrm{tan}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x}-\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{cot}\text{\hspace{0.17em}}x}$

$\frac{1}{1-\mathrm{cos}\text{\hspace{0.17em}}x}-\frac{\mathrm{cos}\text{\hspace{0.17em}}x}{1+\mathrm{cos}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

$\left(\mathrm{sec}\text{\hspace{0.17em}}x+\mathrm{csc}\text{\hspace{0.17em}}x\right)\left(\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\right)-2-\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x$

$\mathrm{tan}\text{\hspace{0.17em}}x$

$-4\mathrm{sec}\text{\hspace{0.17em}}x\mathrm{tan}\text{\hspace{0.17em}}x$

$\mathrm{tan}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{sec}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x$

$±\sqrt{\frac{1}{{\mathrm{cot}}^{2}x}+1}$

$\mathrm{sec}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\frac{±\sqrt{1-{\mathrm{sin}}^{2}x}}{\mathrm{sin}\text{\hspace{0.17em}}x}$

$\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

For the following exercises, verify the identity.

$\mathrm{cos}\text{\hspace{0.17em}}x-{\mathrm{cos}}^{3}x=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}x$

Answers will vary. Sample proof:

$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}x-{\mathrm{cos}}^{3}x& =& \mathrm{cos}\text{\hspace{0.17em}}x\left(1-{\mathrm{cos}}^{2}x\right)\hfill \\ & =& \mathrm{cos}\text{\hspace{0.17em}}x{\mathrm{sin}}^{2}x\hfill \end{array}$

$\mathrm{cos}\text{\hspace{0.17em}}x\left(\mathrm{tan}\text{\hspace{0.17em}}x-\mathrm{sec}\left(-x\right)\right)=\mathrm{sin}\text{\hspace{0.17em}}x-1$

$\frac{1+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{1}{{\mathrm{cos}}^{2}x}+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=1+2\text{\hspace{0.17em}}{\mathrm{tan}}^{2}x$

Answers will vary. Sample proof:
$\frac{1+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{1}{{\mathrm{cos}}^{2}x}+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}={\mathrm{sec}}^{2}x+{\mathrm{tan}}^{2}x={\mathrm{tan}}^{2}x+1+{\mathrm{tan}}^{2}x=1+2{\mathrm{tan}}^{2}x$

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

${\mathrm{cos}}^{2}x-{\mathrm{tan}}^{2}x=2-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x$

Answers will vary. Sample proof:
${\mathrm{cos}}^{2}x-{\mathrm{tan}}^{2}x=1-{\mathrm{sin}}^{2}x-\left({\mathrm{sec}}^{2}x-1\right)=1-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x+1=2-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x$

## Extensions

For the following exercises, prove or disprove the identity.

$\frac{1}{1+\mathrm{cos}\text{\hspace{0.17em}}x}-\frac{1}{1-\mathrm{cos}\left(-x\right)}=-2\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

${\mathrm{csc}}^{2}x\left(1+{\mathrm{sin}}^{2}x\right)={\mathrm{cot}}^{2}x$

False

$\left(\frac{{\mathrm{sec}}^{2}\left(-x\right)-{\mathrm{tan}}^{2}x}{\mathrm{tan}\text{\hspace{0.17em}}x}\right)\left(\frac{2+2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x}{2+2\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x}\right)-2\text{\hspace{0.17em}}{\mathrm{sin}}^{2}x=\mathrm{cos}\text{\hspace{0.17em}}2x$

$\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{\mathrm{sec}\text{\hspace{0.17em}}x}\mathrm{sin}\left(-x\right)={\mathrm{cos}}^{2}x$

False

$\frac{\mathrm{sec}\left(-x\right)}{\mathrm{tan}\text{\hspace{0.17em}}x+\mathrm{cot}\text{\hspace{0.17em}}x}=-\mathrm{sin}\left(-x\right)$

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

Proved with negative and Pythagorean identities

For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression.

$\frac{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }{1-{\mathrm{tan}}^{2}\theta }={\mathrm{sin}}^{2}\theta$

$3\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta +4\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta =3+{\mathrm{cos}}^{2}\theta$

True $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta +4\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta =3\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta +3\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta +{\mathrm{cos}}^{2}\theta =3\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)+{\mathrm{cos}}^{2}\theta =3+{\mathrm{cos}}^{2}\theta$

$\frac{\mathrm{sec}\text{\hspace{0.17em}}\theta +\mathrm{tan}\text{\hspace{0.17em}}\theta }{\mathrm{cot}\text{\hspace{0.17em}}\theta +\mathrm{cos}\text{\hspace{0.17em}}\theta }={\mathrm{sec}}^{2}\theta$

#### Questions & Answers

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