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

7.2 Identities: Cofunction, Double-Angle, and Half-Angle

Use cofunction identities to derive other identities.
Use the double-angle identities to find function values of twice an angle when one function value is known for that angle.
Use the half-angle identities to find function values of half an angle when one function value is known for that angle.
Simplify trigonometric expressions using the double-angle identities and the half-angle identities.

Cofunction Identities

Each of the identities listed below yields a conversion to a cofunction. For this reason, we call them cofunction identities.

We verified the first two of these identities in Section 7.1. The other four can be proved using the first two and the definitions of the trigonometric functions. These six identities hold for all real numbers, and thus, for all angle measures, but if we restrict $θ$ to values such that $0 ° < θ < 90 °$ , or $0 < θ < π / 2$ , then we have a special application to the acute angles of a right triangle.

Comparing graphs can lead to possible identities. On the left below, we see that the graph of $y = sin (x + π / 2)$ is a translation of the graph of $y = sin x$ to the left $π / 2$ units. On the right, we see the graph of $y = cos x$ .

Comparing the graphs, we note a possible identity:

$sin (x + \frac{π}{2}) = cos x .$

The identity can be proved using the identity for the sine of a sum developed in Section 7.1.

Example 1

Prove the identity $sin (x + π / 2) = cos x$ .

Solution

$\begin{matrix} sin (x + \frac{π}{2}) & = & sin x cos \frac{π}{2} + c o s x \sin \frac{π}{2} & Using \sin (u + v) = \sin u \cos v + \cos u \sin v \\ = & \sin x \cdot 0 + \cos x \sin \cdot 1 \\ = & \cos x \end{matrix}$

We now state four more cofunction identities. These new identities that involve the sine and the cosine functions can be verified using previously established identities as seen in Example 1.

Example 2

Find an identity for each of the following.

a) $tan (x + \frac{π}{2})$
b) $sec (x - 90 °)$

Solution

a) We have

$\begin{matrix} tan (x + \frac{π}{2}) & = & \frac{sin (x + \frac{π}{2})}{cos (x + \frac{π}{2})} & Using \tan x = \frac{\sin x}{\cos x} \\ = & \frac{cos x}{- \sin x} & Using confunction identities \\ = & - \cot x . \end{matrix}$

Thus the identity we seek is

$tan (x + \frac{π}{2}) = - cot x .$
b) We have

$sec (x - 90 °) = \frac{1}{cos (x - 90 °)} = \frac{1}{sin x} = csc x .$

Thus, $sec (x - 90 °) = csc x$ .

Now Try Exercises 5 and 7.

Double-Angle Identities

If we double an angle of measure x, the new angle will have measure 2x. Double-angle identities give trigonometric function values of 2x in terms of function values of x. To develop these identities, we will use the sum formulas from the preceding section. We first develop a formula for $sin 2 x$ . Recall that

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

We will consider a number x and substitute it for both u and v in this identity. Doing so gives us

$\begin{matrix} sin (x + x) & = & sin 2 x \\ = & sin x cos x + cos x sin x \\ = & 2 sin x cos x . \end{matrix}$

Our first double-angle identity is thus

$\sin 2 x = 2 \sin x \cos x .$

Double-angle identities for the cosine and the tangent functions can be derived in much the same way as the identity above:

$\begin{matrix} \cos 2 x = \cos^{2} x - \sin^{2} x, & \tan 2 x = \frac{2 \tan x}{1 - \tan^{2} x} . \end{matrix}$

Example 3

Given that $tan θ = - \frac{3}{4}$ and $θ$ is in quadrant II, find each of the following.

a) $sin 2 θ$
b) $cos 2 θ$
c) $tan 2 θ$
d) The quadrant in which $2 θ$ lies

Solution

By drawing a reference triangle as shown, we find that

$sin θ = \frac{3}{5}$

and

$cos θ = - \frac{4}{5} .$

Thus we have the following.

a) $sin 2 θ = 2 sin θ cos θ = 2 \cdot \frac{3}{5} \cdot (- \frac{4}{5}) = - \frac{24}{25}$
b) $\cos 2 θ = \cos^{2} θ - \sin^{2} θ = {(- \frac{4}{5})}^{2} - {(\frac{3}{5})}^{2} = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$
c) $tan 2 θ = \frac{2 tan θ}{1 - {tan}^{2} θ} = \frac{2 \cdot (- \frac{3}{4})}{1 - {(- \frac{3}{4})}^{2}} = \frac{- \frac{3}{2}}{1 - \frac{9}{16}} = - \frac{3}{2} \cdot \frac{16}{7} = - \frac{24}{7}$

Note that $tan 2 θ$ could have been found more easily in this case by simply dividing:

$tan 2 θ = \frac{sin 2 θ}{cos 2 θ} = \frac{- \frac{24}{25}}{\frac{7}{25}} = - \frac{24}{7} .$
d) Since $sin 2 θ$ is negative and $cos 2 θ$ is positive, we know that $2 θ$ is in quadrant IV.

Now Try Exercise 9.

Two other useful identities for $cos 2 x$ can be derived easily, as follows.

$\begin{matrix} \cos 2 x & = & \cos^{2} x - \sin^{2} x & \cos 2 x & = & \cos^{2} x - \sin^{2} x \\ = & (1 - \sin^{2} x) - \sin^{2} x & = & \cos^{2} x - (1 - \cos^{2} x) \\ = & 1 - 2 \sin^{2} x & = & 2 \cos^{2} x - 1 \end{matrix}$

Solving the last two cosine double-angle identities for ${sin}^{2} x$ and ${cos}^{2} x$ , respectively, we obtain two more identities:

$\sin^{2} x = \frac{1 - \cos 2 x}{2} and \cos^{2} x = \frac{1 + \cos 2 x}{2} .$

Using division and these two identities gives us the following useful identity:

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

Example 4

Find an equivalent expression for each of the following.

a) $sin 3 θ$ in terms of function values of $θ$
b) ${cos}^{3} x$ in terms of function values of x or 2x, raised only to the first power

Solution

a) $\begin{matrix} sin 3 θ & = & sin (2 θ + θ) \\ = & sin 2 θ cos θ + cos 2 θ sin θ \\ = & (2 sin θ cos θ) cos θ + (2 {cos}^{2} θ - 1) sin θ & Using sin 2 θ = 2 sin θ cos θ and cos 2 θ = 2 \cos^{2} θ - 1 \\ = & 2 sin θ {cos}^{2} θ + 2 sin θ {cos}^{2} θ - sin θ \\ = & = 4 sin θ {cos}^{2} θ - sin θ \end{matrix}$

We could also substitute ${cos}^{2} θ - s i n^{2} θ$ or $1 - 2 {sin}^{2} θ$ for $cos 2 θ$ . Each substitution leads to a different result, but all results are equivalent.
b) $\begin{matrix} {cos}^{3} x & = & {cos}^{2} x cos x \\ = & \frac{1 + cos 2 x}{2} cos x \\ = & \frac{cos x + cos x cos 2 x}{2} \end{matrix}$

Now Try Exercise 15.

Half-Angle Identities

If we take half of an angle of measure x, the new angle will have measure $x / 2$ . Half-angle identities give trigonometric function values of $x / 2$ in terms of function values of x. To develop these identities, we replace x with $x / 2$ and take square roots. For example,

$\begin{matrix} {sin}^{2} x & = & \frac{1 - cos 2 x}{2} & Solving the identity cos 2 x = 1 - 2 {sin}^{2} x for \sin^{2} x \\ {sin}^{2} \frac{x}{2} & = & \frac{1 - cos 2 \cdot \frac{x}{2}}{2} & Substituting \frac{x}{2} for x \\ {sin}^{2} \frac{x}{2} & = & \frac{1 - cos x}{2} \\ sin \frac{x}{2} & = & \pm \sqrt{\frac{1 - cos x}{2}} . & Taking square roots \end{matrix}$

The formula is called a half-angle formula. The use of $+$ and $-$ depends on the quadrant in which the angle $x / 2$ lies. Half-angle identities for the cosine and the tangent functions can be derived in a similar manner. Two additional formulas for the half-angle tangent identity are listed below.

Example 5

Find $tan (π / 8)$ exactly.

Solution

We have

$\begin{matrix} tan \frac{π}{8} & = & tan \frac{\frac{π}{4}}{2} = \frac{sin \frac{π}{4}}{1 + cos \frac{π}{4}} = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \frac{\frac{\sqrt{2}}{2}}{\frac{2 + \sqrt{2}}{2}} \\ = & \frac{\sqrt{2}}{2 + \sqrt{2}} = \frac{\sqrt{2}}{2 + \sqrt{2}} \cdot \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \sqrt{2} - 1. \end{matrix}$

Now Try Exercise 21.

The identities that we have developed are also useful for simplifying trigonometric expressions.

Example 6

Simplify each of the following.

a) $\frac{sin x cos x}{\frac{1}{2} cos 2 x}$
b) $2 {sin}^{2} \frac{x}{2} + cos x$

Solution

a) We can obtain $2 sin x cos x$ in the numerator by multiplying the expression by $\frac{2}{2}$ :

$\begin{matrix} \frac{sin x cos x}{\frac{1}{2} cos 2 x} = & = & \frac{2}{2} \cdot \frac{sin x cos x}{\frac{1}{2} cos 2 x} = \frac{2 sin x cos x}{cos 2 x} \\ = & \frac{sin 2 x}{cos 2 x} & Using sin 2 x = 2 sin x cos x \\ = & tan 2 x . \end{matrix}$
b) We have

$\begin{matrix} 2 {sin}^{2} \frac{x}{2} + cos x & = & 2 (\frac{1 - cos x}{2}) + cos x & Using sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}}, or {sin}^{2} \frac{x}{2} = \frac{1 - \cos x}{2} \\ = & 1 - \cos x + \cos x \\ = & 1. \end{matrix}$

Now Try Exercise 27.

7.2 Exercise Set

1. Given that $sin (3 π / 10) \approx 0.8090$ and $cos (3 π / 10) \approx 0.5878$ , find each of the following.
1. a) The other four function values for $3 π / 10$
2. b) The six function values for $π / 5$
2. Given that

$sin \frac{π}{12} = \frac{\sqrt{2 - \sqrt{3}}}{2} and cos \frac{π}{12} = \frac{\sqrt{2 + \sqrt{3}}}{2},$

find exact answers for each of the following.
1. a) The other four function values for $π / 12$
2. b) The six function values for $5 π / 12$
3. Given that $sin θ = \frac{1}{3}$ and that the terminal side is in quadrant II, find exact answers for each of the following.
1. a) The other function values for $θ$
2. b) The six function values for $π / 2 - θ$
3. c) The six function values for $θ - π / 2$
4. Given that $cos ϕ = \frac{4}{5}$ and that the terminal side is in quadrant IV, find exact answers for each of the following.
1. a) The other function values for $ϕ$
2. b) The six function values for $π / 2 - ϕ$
3. c) The six function values for $ϕ + π / 2$

Find an equivalent expression for each of the following.

5. $sec (x + \frac{π}{2})$
6. $cot (x - \frac{π}{2})$
7. $tan (x - \frac{π}{2})$
8. $csc (x + \frac{π}{2})$

Find the exact value of $\sin 2 θ$ , $\cos 2 θ$ , $\tan 2 θ$ , and the quadrant in which $2 θ$ , lies.

9. $sin θ = \frac{4}{5}, θ$ in quadrant I
10. $cos θ = \frac{5}{13}, θ$ in quadrant I
11. $cos θ = - \frac{3}{5}, θ$ in quadrant III
12. $tan θ = - \frac{15}{8}, θ$ in quadrant II
13. $tan θ = - \frac{5}{12}, θ$ in quadrant II
14. $sin θ = - \frac{\sqrt{10}}{10}, θ$ in quadrant IV
15. Find an equivalent expression for cos 4x in terms of function values of x.
16. Find an equivalent expression for ${sin}^{4} θ$ in terms of function values of $θ, 2 θ$ , or $4 θ$ , raised only to the first power.

Use the half-angle identities to evaluate exactly.

17. $cos 15 °$
18. $tan 67.5 °$
19. $sin 112.5 °$
20. $cos \frac{π}{8}$
21. $tan 75 °$
22. $sin \frac{5 π}{12}$

Given that $\sin θ = 0.3416$ and $θ$ is in quadrant I, find each of the following using identities.

23. $sin 2 θ$
24. $cos \frac{θ}{2}$
25. $sin \frac{θ}{2}$
26. $sin 4 θ$

Simplify.

27. $2 {cos}^{2} \frac{x}{2} - 1$
28. ${cos}^{4} x - {sin}^{4} x$
29. ${(sin x - cos x)}^{2} + sin 2 x$
30. ${(sin x + cos x)}^{2}$
31. $\frac{2 - {sec}^{2} x}{{sec}^{2} x}$
32. $\frac{1 + sin 2 x + cos 2 x}{1 + sin 2 x - cos 2 x}$
33. ${(- 4 cos x sin x + 2 cos 2 x)}^{2} + {(2 cos 2 x + 4 sin x cos x)}^{2}$
34. $2 sin x {cos}^{3} x - 2 {sin}^{3} x cos x$

Skill Maintenance

Complete the identity. [7.1]

35. $1 - {cos}^{2} x =$
36. ${sec}^{2} x - {tan}^{2} x =$
37. ${sin}^{2} x - 1 =$
38. $1 + {cot}^{2} x =$
39. ${csc}^{2} x - {cot}^{2} x =$
40. $1 + {tan}^{2} x =$
41. $1 - {sin}^{2} x =$
42. ${sec}^{2} x - 1 =$

Consider the following functions (a)–(f). Without graphing them, answer questions 43–46 below. [6.5]

a) $f (x) = 2 sin (\frac{1}{2} x - \frac{π}{2})$
b) $f (x) = \frac{1}{2} cos (2 x - \frac{π}{4}) + 2$
c) $f (x) = - sin [2 (x - \frac{π}{2})] + 2$
d) $f (x) = sin (x + π) - \frac{1}{2}$
e) $f (x) = - 2 cos (4 x - π)$
f) $f (x) = - cos [2 (x - \frac{π}{8})]$

43. Which functions have a graph with an amplitude of 2?
44. Which functions have a graph with a period of $π$ ?
45. Which functions have a graph with a period of $2 π$ ?
46. Which functions have a graph with a phase shift of $π / 4$ ?

Synthesis

47. Find $sin 15 °$ first using a difference identity and then using a half-angle identity. Then compare the results.
48. Given that $cos 51 ° \approx 0.6293$ , find the six function values for $141 °$ .

Simplify.

49. $cos (π - x) + cot x sin (x - \frac{π}{2})$
50. $sin (\frac{π}{2} - x) [sec x - cos x]$
51. $\frac{{cos}^{2} y sin (y + \frac{π}{2})}{{sin}^{2} y sin (\frac{π}{2} - y)}$
52. $\frac{cos x - sin (\frac{π}{2} - x) sin x}{cos x - cos (π - x) tan x}$

Find $sin θ, cos θ$ , and $tan θ$ under the given conditions.

53. $tan \frac{θ}{2} = - \frac{5}{3}, π < θ \leq \frac{3 π}{2}$
54. $cos 2 θ = \frac{7}{12}, \frac{3 π}{2} \leq 2 θ \leq 2 π$
55. Acceleration Due to Gravity. The acceleration due to gravity is often denoted by g in a formula such as $S = \frac{1}{2} g t^{2}$ , where S is the distance that an object falls in time t. The number g relates to motion near the earth’s surface and is generally considered constant. In fact, however, g is not constant, but varies slightly with latitude. If $ϕ$ stands for latitude, in degrees, g is given with good approximation by the formula

$g = 9.78049 (1 + 0.005288 {sin}^{2} ϕ - 0.000006 {sin}^{2} 2 ϕ),$
where g is measured in meters per second per second at sea level.
1. a) Chicago has latitude $42 ° N$ . Find g.
2. b) Philadelphia has latitude $40 ° N$ . Find g.
3. c) Express g in terms of $sin ϕ$ only. That is, eliminate the double angle.
56. Nautical Mile. Latitude is used to measure north–south location on the earth between the equator and the poles. For example, Sydney, Australia has latitude $34 ° S$ . (See the figure.) In Great Britain, the nautical mile is defined as the length of a minute of arc of the earth’s radius. Since the earth is flattened slightly at the poles, a British nautical mile varies with latitude. In fact, it is given, in feet, by the function

$N (ϕ) = 6066 - 31 cos 2 ϕ,$

where $ϕ$ is the latitude in degrees.
1. a) What is the length of a British nautical mile at Sydney?
2. b) What is the length of a British nautical mile at the North Pole?
3. c) Express $N (ϕ)$ in terms of $cos ϕ$ only; that is, do not use the double angle.

..................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 7.2 Identities: Cofunction, Double-Angle, and Half-Angle

Create new playlist

Sign In

Sign Up

Table of Contents for
7.2 Identities: Cofunction, Double-Angle, and Half-Angle