3.1 Solving trigonometric equations with identities (Page 5/9)

Essential precalculus, part 2 Page 5 / 9

Rewriting a trigonometric expression using the difference of squares

Rewrite the trigonometric expression: $4 \cos^{2} θ - 1.$

Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus,

\begin{array}{l} 4 \cos^{2} θ - 1 = {(2 \cos θ)}^{2} - 1 \\ = (2 \cos θ - 1) (2 \cos θ + 1) \end{array}

Try It

Rewrite the trigonometric expression: $25 - 9 \sin^{2} θ .$

This is a difference of squares formula: $25 - 9 \sin^{2} θ = (5 - 3 \sin θ) (5 + 3 \sin θ) .$

Simplify by rewriting and using substitution

Simplify the expression by rewriting and using identities:

\csc^{2} θ - \cot^{2} θ

We can start with the Pythagorean identity.

1 + \cot^{2} θ = \csc^{2} θ

Now we can simplify by substituting $1 + \cot^{2} θ$ for $\csc^{2} θ .$ We have

\begin{array}{l} \csc^{2} θ - \cot^{2} θ = 1 + \cot^{2} θ - \cot^{2} θ \\ = 1 \end{array}

Try It

Use algebraic techniques to verify the identity: $\frac{\cos θ}{1 + \sin θ} = \frac{1 - \sin θ}{\cos θ} .$

(Hint: Multiply the numerator and denominator on the left side by $1 - \sin θ .)$

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

Media

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

Key equations

Pythagorean identities	$\begin{array}{l} \sin^{2} θ + \cos^{2} θ = 1 \\ 1 + \cot^{2} θ = \csc^{2} θ \\ 1 + \tan^{2} θ = \sec^{2} θ \end{array}$
Even-odd identities	$\begin{array}{l} \tan (- θ) = - \tan θ \\ \cot (- θ) = - \cot θ \\ \sin (- θ) = - \sin θ \\ \csc (- θ) = - \csc θ \\ \cos (- θ) = \cos θ \\ \sec (- θ) = \sec θ \end{array}$
Reciprocal identities	$\begin{array}{l} \sin θ = \frac{1}{\csc θ} \\ \cos θ = \frac{1}{\sec θ} \\ \tan θ = \frac{1}{\cot θ} \\ \csc θ = \frac{1}{\sin θ} \\ \sec θ = \frac{1}{\cos θ} \\ \cot θ = \frac{1}{\tan θ} \end{array}$
Quotient identities	$\begin{array}{l} \tan θ = \frac{\sin θ}{\cos θ} \\ \cot θ = \frac{\cos θ}{\sin θ} \end{array}$

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 by simplifying an expression and then verifying 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] .

Section exercises

Verbal

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

All three functions, $F, G,$ and $H,$ are even.

This is because $F (- x) = \sin (- x) \sin (- x) = (- \sin x) (- \sin x) = \sin^{2} x = F (x), G (- x) = \cos (- x) \cos (- x) = \cos x \cos x = \cos^{2} x = G (x)$ and $H (- x) = \tan (- x) \tan (- x) = (- \tan x) (- \tan x) = \tan^{2} x = H (x) .$

Examine the graph of $f (x) = \sec x$ on the interval $[- π, π] .$ How can we tell whether the function is even or odd by only observing the graph of $f (x) = \sec x ?$

After examining the reciprocal identity for $\sec t,$ explain why the function is undefined at certain points.

When $\cos t = 0,$ then $\sec t = \frac{1}{0},$ which is undefined.

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

Algebraic

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

$\sin x \cos x \sec x$

$\sin x$

$\sin (- x) \cos (- x) \csc (- x)$

$\tan x \sin x + \sec x \cos^{2} x$

$\sec x$

$\csc x + \cos x \cot (- x)$

$\frac{\cot t + \tan t}{\sec (- t)}$

$\csc t$

$3 \sin^{3} t \csc t + \cos^{2} t + 2 \cos (- t) \cos t$

$- \tan (- x) \cot (- x)$

$−1$

$\frac{- \sin (- x) \cos x \sec x \csc x \tan x}{\cot x}$

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

$\sec^{2} x$

$(\frac{\tan x}{\csc^{2} x} + \frac{\tan x}{\sec^{2} x}) (\frac{1 + \tan x}{1 + \cot x}) - \frac{1}{\cos^{2} x}$

$\frac{1 - \cos^{2} x}{\tan^{2} x} + 2 \sin^{2} x$

$\sin^{2} x + 1$

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

$\frac{\tan x + \cot x}{\csc x}; \cos x$

$\frac{\sec x + \csc x}{1 + \tan x}; \sin x$

$\frac{1}{\sin x}$

$\frac{\cos x}{1 + \sin x} + \tan x; \cos x$

$\frac{1}{\sin x \cos x} - \cot x; \cot x$

$\frac{1}{\cot x}$

$\frac{1}{1 - \cos x} - \frac{\cos x}{1 + \cos x}; \csc x$

$(\sec x + \csc x) (\sin x + \cos x) - 2 - \cot x; \tan x$

$\tan x$

$\frac{1}{\csc x - \sin x}; \sec x and \tan x$

$\frac{1 - \sin x}{1 + \sin x} - \frac{1 + \sin x}{1 - \sin x}; \sec x and \tan x$

$- 4 \sec x \tan x$

$\tan x; \sec x$

$\sec x; \cot x$

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

$\sec x; \sin x$

$\cot x; \sin x$

$\frac{\pm \sqrt{1 - \sin^{2} x}}{\sin x}$

$\cot x; \csc x$

For the following exercises, verify the identity.

$\cos x - \cos^{3} x = \cos x \sin^{2} x$

Answers will vary. Sample proof:

$\cos x - \cos^{3} x = \cos x (1 - \cos^{2} x)$
$= \cos x \sin^{2} x$

$\cos x (\tan x - \sec (- x)) = \sin x - 1$

$\frac{1 + \sin^{2} x}{\cos^{2} x} = \frac{1}{\cos^{2} x} + \frac{\sin^{2} x}{\cos^{2} x} = 1 + 2 \tan^{2} x$

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

${(\sin x + \cos x)}^{2} = 1 + 2 \sin x \cos x$

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

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

Extensions

For the following exercises, prove or disprove the identity.

$\frac{1}{1 + \cos x} - \frac{1}{1 - \cos (- x)} = - 2 \cot x \csc x$

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

False

$(\frac{\sec^{2} (- x) - \tan^{2} x}{\tan x}) (\frac{2 + 2 \tan x}{2 + 2 \cot x}) - 2 \sin^{2} x = \cos 2 x$

$\frac{\tan x}{\sec x} \sin (- x) = \cos^{2} x$

False

$\frac{\sec (- x)}{\tan x + \cot x} = - \sin (- x)$

$\frac{1 + \sin x}{\cos x} = \frac{\cos x}{1 + \sin (- x)}$

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.

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

$3 \sin^{2} θ + 4 \cos^{2} θ = 3 + \cos^{2} θ$

True $3 \sin^{2} θ + 4 \cos^{2} θ = 3 \sin^{2} θ + 3 \cos^{2} θ + \cos^{2} θ = 3 (\sin^{2} θ + \cos^{2} θ) + \cos^{2} θ = 3 + \cos^{2} θ$

$\frac{\sec θ + \tan θ}{\cot θ + \cos θ} = \sec^{2} θ$

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Source: OpenStax, Essential precalculus, part 2. OpenStax CNX. Aug 20, 2015 Download for free at http://legacy.cnx.org/content/col11845/1.2

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