Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

7.1 Identities: Pythagorean and Sum and Difference

State the Pythagorean identities.
Simplify and manipulate expressions containing trigonometric expressions.
Use the sum and difference identities to find function values.

An identity is an equation that is true for all possible replacements of the variables. The following is a list of the identities studied in Chapter 6.

In this section, we will develop some other important identities.

Pythagorean Identities

We now consider three other identities that are fundamental to a study of trigonometry. They are called the Pythagorean identities. Recall that the equation of a unit circle in the xy-plane is

$x^{2} + y^{2} = 1.$

For any point on the unit circle, the coordinates x and y satisfy this equation. Suppose that a real number s determines a point on the unit circle with coordinates $(x, y)$ , or $(\cos s, \sin s)$ . Then $x = \cos s$ and $y = \sin s$ . Substituting cos s for x and sin s for y in the equation of the unit circle gives us the identity

$\begin{matrix} {(cos s)}^{2} + {(sin s)}^{2} = 1, & Substituting cos s for x and sin s for y \end{matrix}$

which can be expressed as

$\sin^{2} s + \cos^{2} s = 1.$

It is conventional in trigonometry to use the notation $\sin^{2} s$ rather than ${(\sin s)}^{2}$ . Note that $\sin^{2} s \neq \sin s^{2}$ .

The identity $\sin^{2} s + \cos^{2} s = 1$ gives a relationship between the sine and the cosine of any real number s. It is an important Pythagorean identity.

We can divide by ${sin}^{2} s$ on both sides of the preceding identity:

$\begin{matrix} \frac{\sin^{2} s}{\sin^{2} s} + \frac{\cos^{2} s}{\sin^{2} s} = \frac{1}{\sin^{2} s} . & {Dividing by sin}^{2} s \end{matrix}$

Simplifying gives us a second Pythagorean identity:

$1 + \cot^{2} s = \csc^{2} s .$

This equation is true for any replacement of s with a real number for which $\sin^{2} s \neq 0$ , since we divided by $\sin^{2} s$ . But the numbers for which $\sin^{2} s = 0$ $(or sin s = 0)$ are exactly the ones for which the cotangent function and the cosecant function are not defined. Hence our new equation holds for all real numbers s for which cot s and csc s are defined and is thus an identity.

The third Pythagorean identity is obtained by dividing by $\cos^{2} s$ on both sides of the first Pythagorean identity:

$\begin{matrix} \frac{\sin^{2} s}{\cos^{2} s} + \frac{\cos^{2} s}{\cos^{2} s} & = \frac{1}{\cos^{2} s} & Dividing by {cos}^{2} s \\ \tan^{2} s + 1 & = \sec^{2} s . & Simplifying \end{matrix}$

This equation is true for any replacement of s with a real number for which $\cos^{2} s \neq 0$ , since we divided by $\cos^{2} s$ . But the numbers for which $\cos^{2} s = 0$ $(or \cos s = 0)$ are exactly those for which the tangent function and the secant function are not defined. Thus our new equation holds for all real numbers s for which tan s and sec s are defined and is thus an identity.

The identities we have developed hold no matter what symbols are used for the variables. For example, we could write

$\sin^{2} s + \cos^{2} s = 1, \sin^{2} θ + \cos^{2} θ = 1, or \sin^{2} x + \cos^{2} x = 1.$

It is often helpful to express the Pythagorean identities in equivalent forms.

Pythagorean Identities	Equivalent Forms
${sin}^{2} x + {cos}^{2} x = 1$	$\sin^{2} x = 1 - \cos^{2} x$
${sin}^{2} x + {cos}^{2} x = 1$	$\cos^{2} x = 1 - \sin^{2} x$
$1 + \cot^{2} x = \csc^{2} x$	$1 = \csc^{2} x - \cot^{2} x$
$1 + \cot^{2} x = \csc^{2} x$	$\cot^{2} x = \csc^{2} x - 1$
$1 + \tan^{2} x = \sec^{2} x$	$1 = \sec^{2} x - \tan^{2} x$
$1 + \tan^{2} x = \sec^{2} x$	$\tan^{2} x = \sec^{2} x - 1$

Simplifying Trigonometric Expressions

We can factor, simplify, and manipulate trigonometric expressions in the same way that we manipulate strictly algebraic expressions.

Example 1

Multiply and simplify: $\cos x (\tan x - \sec x)$ .

Solution

$\begin{matrix} \cos x & (\tan x - \sec x) \\ = \cos x \tan x - \cos x \sec x & Multiplying \\ = \cos x \frac{\sin x}{\cos x} - \cos x \frac{1}{\cos x} & Recalling the identities \tan x = \frac{\sin x}{\cos x} and \sec x = \frac{1}{\cos x} and substituting \\ = \sin x - 1 & Simplifying \end{matrix}$

Now Try Exercise 3.

There is no general procedure for simplifying trigonometric expressions, but it is often helpful to write everything in terms of sines and cosines, as we did in Example 1. We also look for a Pythagorean identity within a trigonometric expression.

Example 2

Factor and simplify: ${sin}^{2} x {cos}^{2} x + {cos}^{4} x .$

Solution

$\begin{matrix} \sin^{2} x & \cos^{2} x + \cos^{4} x \\ = \cos^{2} x (\sin^{2} x + \cos^{2} x) & Removing a common factor \\ = \cos^{2} x \cdot (1) & Using \sin^{2} x + \cos^{2} x = 1 \\ = \cos^{2} x \end{matrix}$

Now Try Exercise 9 and 13.

Technology Connection

A graphing calculator can be used to perform a partial check of an identity. First, we graph the expression on the left side of the equals sign. Then we graph the expression on the right side using the same screen. If the two graphs are indistinguishable, then we have a partial verification that the equation is an identity. Of course, we can never see the entire graph, so there can always be some doubt. Also, the graphs may not overlap precisely, but you may not be able to tell because the difference between the graphs may be less than the width of a pixel. However, if the graphs are obviously different, we know that a mistake has been made.

Consider the identity in Example 1:

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

Recalling that $sec x = 1 / cos x$ , we enter

$y_{1} = \cos x (\tan x - 1 / \cos x) and y_{2} = \sin x - 1.$

To graph, we first select SEQUENTIAL mode. Then we select the “line”-graph style for $y_{1}$ and the “path”-graph style, denoted by $-○$ , for $y_{2}$ . The calculator will graph $y_{1}$ first. Then it will graph $y_{2}$ as the circular cursor traces the leading edge of the graph, allowing us to determine whether the graphs coincide. As you can see in the graph screen at left, the graphs appear to be identical. Thus, $\cos x (\tan x - \sec x) = \sin x - 1$ is most likely an identity.

The TABLE feature can also be used to check identities. Note in the table at left that the function values are the same except for those values of x for which $\cos x = 0$ . The domain of $y_{1}$ excludes these values. The domain of $y_{2}$ is the set of all real numbers. Thus all real numbers except $\pm π / 2, \pm 3 π / 2$ , $\pm 5 π / 2, \dots$ are possible replacements for x in the identity. Recall that an identity is an equation that is true for all possible replacements.

Example 3

Simplify each of the following trigonometric expressions.

a) $\frac{\cot (- θ)}{\csc (- θ)}$
b) $\frac{2 \sin^{2} t + \sin t - 3}{1 - \cos^{2} t - \sin t}$

Solution

a) $\begin{matrix} \frac{\cot (- θ)}{\csc (- θ)} & = \frac{\frac{\cos (- θ)}{\sin (- θ)}}{\frac{1}{\sin (- θ)}} & Rewriting in terms of sines and cosines \\ = \frac{\cos (- θ)}{\sin (- θ)} \cdot \sin (- θ) & Multiplying by the reciprocal, \sin (- θ) / 1 \\ = \cos (- θ) & Removing a factor of 1, \sin (- θ) / \sin (- θ) \\ = \cos θ & The cosine function is even. \end{matrix}$

Recall that the sine function is odd, $\sin (- θ) = - \sin θ$ , and the cosine function is even, $\cos (- θ) = \cos θ$ . It can be shown that the tangent, the cotangent, and the cosecant functions are odd and the secant function is even. We can also simplify this expression using those identities:

$\begin{matrix} \frac{\cot (- θ)}{\csc (- θ)} = \frac{- \cot θ}{- \csc θ} = \frac{\cot θ}{\csc θ} & = \frac{\frac{\cos θ}{\sin θ}}{\frac{1}{\sin θ}} \\ = \frac{\cos θ}{\sin θ} \cdot \frac{\sin θ}{1} = \cos θ . \end{matrix}$
b) $\frac{2 {sin}^{2} t + sin t - 3}{1 - {cos}^{2} t - sin t}$

$\begin{matrix} = \frac{2 \sin^{2} t + \sin t - 3}{\sin^{2} t - \sin t} \\ = \frac{(2 \sin t + 3) (\sin t - 1)}{\sin t (\sin t - 1)} & Substituting \sin^{2} t for 1 - \cos^{2} t \\ = \frac{2 \sin t + 3}{\sin t} & Factoring in both the numerator and the denominator \\ = \frac{2 \sin t}{\sin t} + \frac{3}{\sin t} & Simplifying \\ = 2 + \frac{3}{\sin t}, or 2 + 3 \csc t \end{matrix}$

Now Try Exercises 17 and 19.

We can add and subtract trigonometric rational expressions in the same way that we do algebraic expressions, writing expressions with a common denominator before adding and subtracting numerators.

Example 4

Add and simplify: $\frac{\cos x}{1 + \sin x} + \tan x$ .

Solution

$\begin{matrix} \frac{\cos x}{1 + \sin x} + \tan x & = \frac{\cos x}{1 + \sin x} + \frac{\sin x}{\cos x} & Using \tan x = \frac{\sin x}{\cos x} \\ = \frac{\cos x}{1 + \sin x} \cdot \frac{\cos x}{\cos x} + \frac{\sin x}{\cos x} \cdot \frac{1 + \sin x}{1 + \sin x} & Multiplying by forms of 1 \\ = \frac{\cos^{2} x + \sin x + \sin^{2} x}{\cos x (1 + \sin x)} & Adding \\ = \frac{1 + \sin x}{\cos x (1 + \sin x)} & Using \sin^{2} x + \cos^{2} x = 1 \\ = \frac{1}{\cos x}, or \sec x & Simplifying \end{matrix}$

Now Try Exercise 27.

When radicals occur, the use of absolute value is sometimes necessary, but it can be difficult to determine when to use it. In Examples 5 and 6, we will assume that all radicands are nonnegative. This means that the identities are meant to be confined to certain quadrants.

Example 5

Multiply and simplify: $\sqrt{{sin}^{3} x cos x} \cdot \sqrt{cos x}$ .

Solution

$\begin{matrix} \sqrt{\sin^{3} x \cos x} \cdot \sqrt{\cos x} & = \sqrt{\sin^{3} x \cos^{2} x} \\ = \sqrt{\sin^{2} x \cos^{2} x \sin x} \\ = \sin x \cos x \sqrt{\sin x} \end{matrix}$

Now Try Exercise 31.

Example 6

Rationalize the denominator: $\sqrt{\frac{2}{\tan x}}$ .

Solution

$\begin{matrix} \sqrt{\frac{2}{\tan x}} & = \sqrt{\frac{2}{\tan x} \cdot \frac{\tan x}{\tan x}} \\ = \sqrt{\frac{2 \tan x}{\tan^{2} x}} \\ = \frac{\sqrt{2 \tan x}}{\tan x} \end{matrix}$

Now Try Exercise 37.

Often in calculus, a substitution is a useful manipulation, as we show in the following example.

Example 7

Express $\sqrt{9 + x^{2}}$ as a trigonometric function of $θ$ without using radicals by letting $x = 3 \tan θ$ . Assume that $0 < θ < π / 2$ . Then find $\sin θ$ and $\cos θ$ .

Solution

We have

$\begin{matrix} \sqrt{9 + x^{2}} & = \sqrt{9 + {(3 \tan θ)}^{2}} & Substituting 3 tan θ for x \\ = \sqrt{9 + 9 \tan^{2} θ} \\ = \sqrt{9 (1 + \tan^{2} θ)} & Factoring \\ = \sqrt{9 \sec^{2} θ} & Using 1 + {tan}^{2} x = {sec}^{2} x \\ = 3 | \sec θ | = 3 \sec θ . & For 0 < θ < π /2, sec θ > 0, so | sec θ | = sec θ . \\ We can express & \sqrt{9 + x^{2}} = 3 \sec θ as \\ \sec θ & = \frac{\sqrt{9 + x^{2}}}{3} . \end{matrix}$

In a right triangle, we know that $\sec θ$ is hypotenuse/adjacent, when $θ$ is one of the acute angles. Using the Pythagorean theorem, we can determine that the side opposite $θ$ is x. Then from the right triangle, we see that

$\sin θ = \frac{x}{\sqrt{9 + x^{2}}} and \cos θ = \frac{3}{\sqrt{9 + x^{2}}} .$

Now Try Exercise 45.

Sum and Difference Identities

We now develop some important identities involving sums or differences of two numbers (or angles), beginning with an identity for the cosine of the difference of two numbers. We use the letters u and v for these numbers.

Let’s consider a real number u in the interval $[π / 2, π]$ and a real number v in the interval $[0, π / 2]$ . These determine points A and B on the unit circle, as shown below. The arc length s is $u - v$ , and we know that $0 \leq s \leq π$ . Recall that the coordinates of A are $(\cos u, \sin u)$ , and the coordinates of B are $(\cos v, \sin v)$ .

Using the distance formula, we can write an expression for the distance AB:

$A B = \sqrt{{(cos u - cos v)}^{2} + {(sin u - sin v)}^{2}} .$

This can be simplified as follows:

$\begin{matrix} A B & = \sqrt{\cos^{2} u - 2 \cos u \cos v + \cos^{2} v + \sin^{2} u - 2 \sin u \sin v + \sin^{2} v} \\ = \sqrt{(\sin^{2} u + \cos^{2} u) + (\sin^{2} v + \cos^{2} v) - 2 (\cos u \cos v + \sin u \sin v)} \\ = \sqrt{2 - 2 (\cos u \cos v + \sin u \sin v)} . \end{matrix}$

Now let’s imagine rotating the circle so that point B is at (1, 0), as shown at left. Although the coordinates of point A are now (cos s, sin s), the distance AB has not changed.

Again, we use the distance formula to write an expression for the distance AB:

$A B = \sqrt{{(cos s - 1)}^{2} + {(sin s - 0)}^{2}} .$

This can be simplified as follows:

$\begin{matrix} A B & = \sqrt{\cos^{2} s - 2 \cos s + 1 + \sin^{2} s} \\ = \sqrt{(\sin^{2} s + \cos^{2} s) + 1 - 2 \cos s} \\ = \sqrt{2 - 2 \cos s} . \end{matrix}$

Equating our two expressions for AB, we obtain

$\sqrt{2 - 2 (\cos u \cos v + \sin u \sin v)} = \sqrt{2 - 2 \cos s} .$

Solving this equation for cos s gives

(1)

$\cos s = \cos u \cos v + \sin u \sin v .$

But $s = u - v$ , so we have the equation

(2)

$\cos (u - v) = \cos u \cos v + \sin u \sin v .$

Formula (1) above holds when s is the length of the shortest arc from A to B. Given any real numbers u and v, the length of the shortest arc from A to B is not always $u - v$ . In fact, it could be $v - u$ . However, since $\cos (- x) = \cos x$ , we know that $\cos (v - u) = \cos (u - v)$ . Thus, cos s is always equal to $\cos (u - v)$ . Formula (2) holds for all real numbers u and v. That formula is thus the identity we sought:

$\cos (u - v) = \cos u \cos v + \sin u \sin v .$

The cosine sum formula follows easily from the one we have just derived. Let’s consider $\cos (u + v)$ . This is equal to $\cos [u - (- v)]$ , and by the identity above, we have

$\begin{matrix} \cos (u + v) & = \cos [u - (- v)] \\ = \cos u \cos (- v) + \sin u \sin (- v) . \end{matrix}$

But $\cos (- v) = \cos v$ and $\sin (- v) = - \sin v$ , so the identity we seek is the following:

$\cos (u + v) = \cos u \cos v - \sin u \sin v .$

Example 8

Find $\cos (5 π / 12)$ exactly.

Solution

We can express $5 π / 12$ as a difference of two numbers whose exact sine and cosine values are known:

$\frac{5 π}{12} = \frac{9 π}{12} - \frac{4 π}{12}, or \frac{3 π}{4} - \frac{π}{3} .$

Then, using $\cos (u - v) = \cos u \cos v + \sin u \sin v$ , we have

$\begin{matrix} \cos \frac{5 π}{12} = \cos (\frac{3 π}{4} - \frac{π}{3}) & = \cos \frac{3 π}{4} \cos \frac{π}{3} + \sin \frac{3 π}{4} \sin \frac{π}{3} \\ = - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} \\ = - \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} \\ = \frac{\sqrt{6} - \sqrt{2}}{4} . \end{matrix}$

Now Try Exercise 51.

Consider $\cos (π / 2 - θ)$ . We can use the identity for the cosine of a difference to simplify as follows:

$\begin{matrix} \cos (\frac{π}{2} - θ) & = \cos \frac{π}{2} \cos θ + \sin \frac{π}{2} \sin θ \\ = 0 \cdot \cos θ + 1 \cdot \sin θ \\ = \sin θ . \end{matrix}$

Thus we have developed the identity

(3)

$\sin θ = \cos (\frac{π}{2} - θ) . This cofunction identity first appeared in Section 6.1.$

This identity holds for any real number $θ$ . From it we can obtain an identity for the cosine function. We first let $α$ be any real number. Then we replace $θ$ in $\sin θ = \cos (π / 2 - θ)$ with $π / 2 - α$ . This gives us

$\sin (\frac{π}{2} - α) = \cos [\frac{π}{2} - (\frac{π}{2} - α)] = \cos α,$

which yields the identity

(4)

$\cos α = \sin (\frac{π}{2} - α) .$

Using identities (3) and (4) and the identity for the cosine of a difference, we can obtain an identity for the sine of a sum. We start with identity (3) and substitute $u + v$ for $θ :$

$\begin{matrix} \sin θ & = \cos (\frac{π}{2} - θ) & Identity (3) \\ \sin (u + v) & = \cos [\frac{π}{2} - (u + v)] & Substituting u + v for θ \\ = \cos [(\frac{π}{2} - u) - v] \\ = \cos (\frac{π}{2} - u) \cos v + \sin (\frac{π}{2} - u) \sin v & Using the identity for the cosine of a difference \\ = \sin u \cos v + \cos u \sin v . & Using identities (3) and (4) \end{matrix}$

Thus the identity we seek is

$\sin (u + v) = \sin u \cos v + \cos u \sin v .$

To find a formula for the sine of a difference, we can use the identity just derived, substituting $- v$ for v:

$\sin (u + (- v)) = \sin u \cos (- v) + \cos u \sin (- v) .$

Simplifying gives us

$\sin (u - v) = \sin u \cos v - \cos u \sin v .$

Example 9

Find sin $105 °$ exactly.

Solution

We express $105 °$ as the sum of two measures:

$105 ° = 45 ° + 60 ° .$

Then

$\begin{matrix} \sin 105 ° & = \sin (45 ° + 60 °) \\ = \sin 45 ° \cos 60 ° + \cos 45 ° \sin 60 ° & Using \sin (u + v) = \sin u \cos v + \cos u \sin v \\ = \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} \\ = \frac{\sqrt{2} + \sqrt{6}}{4} . \end{matrix}$

Now Try Exercise 55.

Formulas for the tangent of a sum or a difference can be derived using identities already established. A summary of the sum and difference identities follows.

Example 10

Find tan $15 °$ exactly.

Solution

We rewrite $15 °$ as $45 ° - 30 °$ and use the identity for the tangent of a difference:

$\begin{matrix} tan 15 ° & = & tan (45 ° - 30 °) \\ = & \frac{tan 45 ° - tan 30 °}{1 + tan 45 ° tan 30 °} \\ = & \frac{1 - \sqrt{3} / 3}{1 + 1 \cdot \sqrt{3} / 3} \\ = & \frac{3 - \sqrt{3}}{3 + \sqrt{3}} . \end{matrix}$

Now Try Exercise 53.

Example 11

Assume that $sin α = \frac{2}{3}$ and $sin β = \frac{1}{3}$ and that $α$ and $β$ are between 0 and $π / 2$ . Then evaluate $sin (α + β)$ .

Solution

Using the identity for the sine of a sum, we have

$\begin{matrix} sin (α + β) & = & sin α cos β + cos α sin β \\ = & \frac{2}{3} cos β + \frac{1}{3} cos α . \end{matrix}$

To finish, we need to know the values of $cos β$ and $cos α$ . Using reference triangles and the Pythagorean theorem, we can determine these values from the diagrams:

$\begin{matrix} cos α = \frac{\sqrt{5}}{3} & and & cos β = \frac{2 \sqrt{2}}{3} . & Cosine values are positive in the first quadrant. \end{matrix}$

Substituting these values gives us

$\begin{matrix} sin (α + β) & = & \frac{2}{3} \cdot \frac{2 \sqrt{2}}{3} + \frac{1}{3} \cdot \frac{\sqrt{5}}{3} \\ = & \frac{4}{9} \sqrt{2} + \frac{1}{9} \sqrt{5}, or \frac{4 \sqrt{2} + \sqrt{5}}{9} . \end{matrix}$

Now Try Exercise 65.

Example 12

Assume that $cos α = - \frac{4}{5}$ with $α$ between $π$ and $3 π / 2$ and that $cos β = - \frac{2}{5}$ with $β$ between $π / 2$ and $π$ . Then evaluate $cos (α - β)$ .

Solution

Using the identity for the cosine of a difference, we have

$\begin{matrix} cos (α - β) & = & cos α cos β + sin α sin β \\ = & (- \frac{4}{5}) \cdot (- \frac{2}{5}) + sin α sin β \\ = & \frac{8}{25} + sin α sin β . \end{matrix}$

We need to know the values of $sin α$ and $sin β$ . Using reference triangles and the Pythagorean theorem, we can determine these values from the diagrams:

$\begin{matrix} sin α = - \frac{3}{5} & and & sin β = \frac{\sqrt{21}}{5} . \end{matrix}$

Substituting these values gives us

$\begin{matrix} cos (α - β) & = & \frac{8}{25} + (- \frac{3}{5}) \cdot (\frac{\sqrt{21}}{5}) \\ = & \frac{8}{25} - \frac{3 \sqrt{21}}{25} = \frac{8 - 3 \sqrt{21}}{25} . \end{matrix}$

Now Try Exercise 69.

7.1 Exercise Set

Multiply and simplify.

1. $(sin x - cos x) (sin x + cos x)$
2. $tan x (cos x - csc x)$
3. $cos y sin y (sec y + csc y)$
4. $(sin x + cos x) (sec x + csc x)$
5. ${(sin ϕ - cos ϕ)}^{2}$
6. ${(1 + tan x)}^{2}$
7. $(sin x + csc x) ({sin}^{2} x + {csc}^{2} x - 1)$
8. $(1 - sin t) (1 + sin t)$

Factor and simplify.

9. $sin x cos x + {cos}^{2} x$
10. ${tan}^{2} θ - {cot}^{2} θ$
11. ${sin}^{4} x - {cos}^{4} x$
12. $4 {sin}^{2} y + 8 sin y + 4$
13. $2 {cos}^{2} x + cos x - 3$
14. $3 {cot}^{2} β + 6 cot β + 3$
15. ${sin}^{3} x + 27$
16. $1 - 125 {tan}^{3} s$

Simplify.

17. $\frac{{sin}^{2} x cos x}{{cos}^{2} x sin x}$
18. $\frac{30 {sin}^{3} x cos x}{6 {cos}^{2} x sin x}$
19. $\frac{{sin}^{2} x + 2 sin x + 1}{sin x + 1}$
20. $\frac{{cos}^{2} α - 1}{cos α + 1}$
21. $\frac{4 tan t sec t + 2 sec t}{6 tan t sec t + 2 sec t}$
22. $\frac{csc (- x)}{cot (- x)}$
23. $\frac{{sin}^{4} x - {cos}^{4} x}{{sin}^{2} x - {cos}^{2} x}$
24. $\frac{4 {cos}^{3} x}{{sin}^{2} x} \cdot {(\frac{sin x}{4 cos x})}^{2}$
25. $\frac{5 cos ϕ}{{sin}^{2} ϕ} \cdot \frac{{sin}^{2} ϕ - sin ϕ c o s ϕ}{{sin}^{2} ϕ - {cos}^{2} ϕ}$
26. $\frac{{tan}^{2} y}{sec y} \div \frac{3 {tan}^{3} y}{sec y}$
27. $\frac{1}{{sin}^{2} s - {cos}^{2} s} - \frac{2}{cos s - sin s}$
28. ${(\frac{sin x}{cos x})}^{2} - \frac{1}{{cos}^{2} x}$
29. $\frac{{sin}^{2} θ - 9}{2 cos θ + 1} \cdot \frac{10 cos θ + 5}{3 sin θ + 9}$
30. $\frac{9 {cos}^{2} α - 25}{2 cos α - 2} \cdot \frac{{cos}^{2} α - 1}{6 cos α - 10}$

Simplify. Assume that all radicands are nonnegative.

31. $\sqrt{{sin}^{2} x cos x} \cdot \sqrt{cos x}$
32. $\sqrt{{cos}^{2} x sin x} \cdot \sqrt{sin x}$
33. $\sqrt{cos α {sin}^{2} α} - \sqrt{{cos}^{3} α}$
34. $\sqrt{{tan}^{2} x - 2 tan x sin x + {sin}^{2} x}$
35. $(1 - \sqrt{sin y}) (\sqrt{sin y} + 1)$
36. $\sqrt{cos θ} (\sqrt{2 cos θ} + \sqrt{sin θ cos θ})$

Rationalize the denominator.

37. $\sqrt{\frac{sin x}{cos x}}$
38. $\sqrt{\frac{cos x}{tan x}}$
39. $\sqrt{\frac{{cos}^{2} y}{2 {sin}^{2} y}}$
40. $\sqrt{\frac{1 - cos β}{1 + cos β}}$

Rationalize the numerator.

41. $\sqrt{\frac{cos x}{sin x}}$
42. $\sqrt{\frac{sin x}{cot x}}$
43. $\sqrt{\frac{1 + sin y}{1 - sin y}}$
44. $\sqrt{\frac{{cos}^{2} x}{2 {sin}^{2} x}}$

Use the given substitution to express the given radical expression as a trigonometric function without radicals. Assume that $a > 0$ and $0 < θ < π / 2$ . Then find expressions for the indicated trigonometric functions.

45. Let $x = a sin θ$ in $\sqrt{a^{2} - x^{2}}$ . Then find $cos θ$ and $tan θ$ .
46. Let $x = 2 tan θ$ in $\sqrt{4 + x^{2}}$ . Then find $sin θ$ and $cos θ$ .
47. Let $x = 3 sec θ$ in $\sqrt{x^{2} - 9}$ . Then find $sin θ$ and $cos θ$ .
48. Let $x = a sec θ$ in $\sqrt{x^{2} - a^{2}}$ . Then find $sin θ$ and $cos θ$ .

Use the given substitution to express the given radical expression as a trigonometric function without radicals. Assume that $0 < θ < π / 2$ .

49. Let $x = sin θ$ in $\frac{x^{2}}{\sqrt{1 - x^{2}}}$ .
50. Let $x = 4 \sec θ$ in $\frac{\sqrt{x^{2} - 16}}{x^{2}}$ .

Use the sum and difference identities to evaluate exactly.

51. $sin \frac{π}{12}$
52. $cos 75 °$
53. $tan 105 °$
54. $tan \frac{5 π}{12}$
55. $cos 15 °$
56. $sin \frac{7 π}{12}$

First write each of the following as a trigonometric function of a single angle. Then evaluate.

57. $sin 37 ° cos 22 ° + cos 37 ° sin 22 °$
58. $cos 83 ° cos 53 ° + sin 83 ° sin 53 °$
59. $cos 19 ° cos 5 ° - sin 19 ° sin 5 °$
60. $sin 40 ° cos 15 ° - cos 40 ° sin 15 °$
61. $\frac{\tan 20 ° + \tan 32 °}{1 - \tan 20 ° \tan 32 °}$
62. $\frac{tan 35 ° - \tan 12 °}{1 + tan 35 ° tan 12 °}$
63. Derive the formula for the tangent of a sum.
64. Derive the formula for the tangent of a difference.

Assuming that $sin u = \frac{3}{5}$ and $sin v = \frac{4}{5}$ and that u and v are between 0 and $π / 2$ , evaluate each of the following exactly.

65. $cos (u + v)$
66. $tan (u - v)$
67. $sin (u - v)$
68. $cos (u - v)$

Assuming that $cos α = - \frac{3}{7}$ with $α$ between $π / 2$ and $π$ and that $cos β = \frac{8}{9}$ with $β$ between $3 π / 2$ and $2 π$ , evaluate each of the following exactly.

69. $cos (α + β)$
70. $sin (α - β)$

Assuming that $sin θ = 0.6249$ and $cos ϕ = 0.1102$ and that both $θ$ and $ϕ$ are first-quadrant angles, evaluate each of the following.

71. $tan (θ + ϕ)$
72. $sin (θ - ϕ)$
73. $cos (θ - ϕ)$
74. $cos (θ + ϕ)$

Simplify.

75. $sin (α + β) + s i n (α - β)$
76. $cos (α + β) - c o s (α - β)$
77. $cos (u + v) cos v + sin (u + v) sin v$
78. $sin (u - v) cos v + cos (u - v) sin v$

Skill Maintenance

Solve. [1.5]

79. $2 x - 3 = 2 (x - \frac{3}{2})$
80. $x - 7 = x + 3.4$

Given that $sin 31 ° = 0.5150$ and $cos 31 ° = 0.8572$ , find the specified function value. [6.1]

81. $sec 59 °$
82. $tan 59 °$

Synthesis

Angles Between Lines. One of the identities gives an easy way to find an angle formed by two lines. Consider two lines with equations $l_{1}$ : $y = m_{1} x + b_{1}$ and $l_{2}$ : $y = m_{2} x + b_{2}$ .

The slopes $m_{1}$ and $m_{2}$ are the tangents of the angles $θ_{1}$ and $θ_{2}$ that the lines form with the positive direction of the x-axis. Thus we have $m_{1} = \tan θ_{1}$ and $m_{2} = tan θ_{2}$ . To find the measure of $θ_{2} - θ_{1}$ , or $ϕ$ , we proceed as follows:

$\begin{matrix} tan ϕ & = & tan (θ_{2} - θ_{1}) \\ = & \frac{tan θ_{2} - tan θ_{1}}{1 + tan θ_{2} tan θ_{1}} \\ = & \frac{m_{2} - m_{1}}{1 + m_{2} m_{1}} . \end{matrix}$

This formula also holds when the lines are taken in the reverse order. When $ϕ$ is acute, $tan ϕ$ will be positive. When $ϕ$ is obtuse, $tan ϕ$ will be negative.

Find the measure of the angle from $l_{1}$ to $l_{2}$ .

83. $l_{1} : 2 x = 3 - 2 y$ ,

$l_{2} : x + y = 5$
84. $l_{1} : 3 y = \sqrt{3} x + 3$ ,

$l_{2} : y = \sqrt{3} x + 2$
85. $l_{1} : y = 3$ ,

$l_{2} : x + y = 5$
86. $l_{1} : 2 x + y - 4 = 0$ ,

$l_{2} : y - 2 x + 5 = 0$
87. Rope Course and Climbing Wall. For a rope course and climbing wall, a guy wire R is attached 47 ft high on a vertical pole. Another guy wire S is attached 40 ft above the ground on the same pole. (Source: Experiential Resources, Inc., Todd Domeck, Owner) Find the angle $α$ between the wires if they are attached to the ground 50 ft from the pole.
88. Circus Guy Wire. In a circus, a guy wire A is attached to the top of a 30-ft pole. Wire B is used for performers to walk up to the tight wire, 10 ft above the ground. Find the angle $ϕ$ between the wires if they are attached to the ground 40 ft from the pole.
89. Given that $f (x) = cos x$ , show that

$\frac{f (x + h) - f (x)}{h} = cos x (\frac{cos h - 1}{h}) - sin x (\frac{sin h}{h}) .$
90. Given that $f (x) = sin x$ , show that

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

Show that each of the following is not an identity by finding a replacement or replacements for which the sides of the equation do not name the same number.

91. $\frac{sin 5 x}{x} = sin 5$
92. $\sqrt{{sin}^{2} θ} = sin θ$
93. $cos (2 α) = 2 cos α$
94. $sin (- x) = sin x$
95. $\frac{cos 6 x}{cos x} = 6$
96. ${tan}^{2} θ + {cot}^{2} θ = 1$

Find the slope of line $l_{1}$ , where $m_{2}$ is the slope of line $l_{2}$ and $ϕ$ is the smallest positive angle from $l_{1}$ to $l_{2}$ .

97. $m_{2} = \frac{2}{3}, ϕ = 30 °$
98. $m_{2} = \frac{4}{3}, ϕ = 45 °$
99. Line $l_{1}$ contains the points $(- 3, 7)$ and $(- 3, - 2)$ . Line $l_{2}$ contains $(0, - 4)$ and $(2, 6)$ . Find the smallest positive angle from $l_{1}$ to $l_{2}$ .
100. Line $l_{1}$ contains the points $(- 2, 4)$ and $(5, - 1)$ . Find the slope of line $l_{2}$ such that the angle from $l_{1}$ to $l_{2}$ is $45 °$ .
101. Find an identity for $cos 2 θ$ . (Hint: $2 θ = θ + θ$ .)
102. Find an identity for $sin 2 θ$ . (Hint: $2 θ = θ + θ$ .)

Derive the identity.

103. $tan (x + \frac{π}{4}) = \frac{1 + tan x}{1 - tan x}$
104. $sin (x - \frac{3 π}{2}) = cos x$
105. $sin (α + β) + s i n (α - β) = 2 sin α cos β$
106. $\frac{sin (α + β)}{cos (α - β)} = \frac{tan α + t a n β}{1 + tan α tan β}$

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for 7.1 Identities: Pythagorean and Sum and Difference

Create new playlist

Sign In

Sign Up

Table of Contents for
7.1 Identities: Pythagorean and Sum and Difference