Trigonometric Functions of Real Numbers

8/11/2019 Trigonometric Functions of Real Numbers

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Section 5.2 Trigonometric Functions of Real Numbers

The Trigonometric Functions

EXAMPLE: Use the Table below to find the six trigonometric functions of each given realnumber t.

(a) t=

3 (b) t=

2

1


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EXAMPLE: Use the Table below to find the six trigonometric functions of each given realnumber t.

(a) t=

3 (b) t=

2

Solution:

(a) From the Table, we see that the terminal point determined byt = /3 is P(1/2,

3/2). Since the coordinates are x = 1/2 and

y=

3/2, we have

sin3

=3

2 cos

3 =1

2 tan

3 =

3/21/2

=

3

csc

3 =

2

3

3 sec

3 = 2 cot

3 =

1/23/2

=

3

3

(b) The terminal point determined by /2 is P(0, 1). So

sin

2 = 1 cos

2 = 0 csc

2 =

1

1= 1 cot

2 =

0

1= 0

But tan /2 and sec /2 are undefined because x = 0 appears in the denominator in each oftheir definitions.

EXAMPLE: Find the six trigonometric functions of each given real number t =

4.

Solution: From the Table above, we see that the terminal point determined by t = /4 isP(

2/2,

2/2). Since the coordinates are x=

2/2 and y =

2/2,we have

sin

4 =

2

2 cos

4 =

2

2 tan

4 =

2/22/2

= 1

csc

4 =

2 sec

4 =

2 cot

4 =

2/22/2

= 1

2


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Values of the Trigonometric Functions

EXAMPLE:

(a) cos

3 >0,because the terminal point oft =

3is in Quadrant I.

(b) tan 4>0, because the terminal point oft = 4 is in Quadrant III.

(c) If cos t 0,then the terminal point oft must be in Quadrant II.

EXAMPLE: Determine the sign of each function.

(a) cos74

(b) tan 1

Solution:

(a) Positive (b) Positive

EXAMPLE: Find each value.

(a) cos2

3 (b) tan

3

(c) sin

19

4

3


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(a) cos2

3 (b) tan

3

(c) sin

19

4

Solution:

(a) Since2

3 =

3 3

=3

3

3 =

3

the reference number for 2/3 is /3 (see Figure (a) below) and the terminal point of 2/3 is

in Quadrant II. Thus cos(2/3) is negative and

(b) The reference number for/3 is /3 (see Figure (b) below). Since the terminal point of/3 is in Quadrant IV, tan(/3) is negative. Thus

(c) Since19

4 =

20 4

=20

4

4 = 5

4

the reference number for 19/4 is /4 (see Figure (c) below) and the terminal point of 19/4is in Quadrant II. Thus sin(19/4) is positive and


(a) sin2

3 (b) tan

4

3

(c) cos

14

3

4


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(a) sin2

3 (b) tan

4

3

(c) cos

14

3

Solution:

(a) Since2

3 =

3 3

=3

3

3 =

3

the reference number for 2/3 is /3 and the terminal point of 2/3 is in Quadrant II. Thussin(2/3) is positive and

sin2

3 = sin

3 =

3

2

(b) Since

43

= 3+3

= 33

3 =

3

the reference number for 4/3 is/3 and the terminal point of4/3 is in Quadrant II. Thustan(4/3) is negative and

tan

4

3

= tan

3

= 3(c) Since

14

3 =

15 3

=15

3

3 = 5

3

the reference number for 14/3 is/3 and the terminal point of 14/3 is in Quadrant II. Thuscos(14/4) is negative and

cos14

3 = cos

3 = 1

2

EXAMPLE: Evaluate

(a) sin3

(b) cos76

(c) tan114

(d) sec173

(e) csc172

(f) cot1216

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EXAMPLE: Evaluate

(a) sin

3 (b) cos

7

6 (c) tan

11

4 (d) sec

17

3 (e) csc

17

2 (f) cot

121

6

Solution:

(a) The reference number for /3 is /3. Since the terminal point of /3 is in Quadrant I,sin(/3) is positive. Thus

sin

3= sin

3 =

3

2

(b) Since 76

= 6+6

= 66

+ 6

= + 6

, the reference number for 7/6 is /6 and the terminalpoint of 7/6 is in Quadrant III. Thus cos(7/6) is negative and

cos7

6 = cos

6 =

3

2

(c) Since 114

= 124

= 124

4 = 3

4, the reference number for 11/4 is /4 and the

terminal point of 11/6 is in Quadrant II. Thus tan(11/4) is negative and

tan11

4 = tan

4 = 1

(d) Since 173

= 183

= 183

3 = 6

3, the reference number for 17/3 is /3 and the

terminal point of 17/3 is in Quadrant IV. Thus sec(17/3) is positive and

sec17

3 = sec

3 = 2

(e) Since 172

= 16+2

= 162

+ 2

= 8+ 2

, the reference number for 17/2 is /2 and theterminal point of 17/2 is in Quadrant I (II). Thus csc(17/2) is positive and

csc17

2 = csc

2 = 1

(f) Since 1216

= 120+6

= 1206

+ 6

= 20+ 6

, the reference number for 121/6 is /6 and theterminal point of 121/6 is in Quadrant I. Thus cot(121/6) is positive and

cot121

6 = cot

6 =

3

EXAMPLE: Use the even-odd properties of the trigonometric functions to determine each value.

(a) sin

6

(b) cos

4

(c) csc

3

6


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EXAMPLE: Use the even-odd properties of the trigonometric functions to determine each value.

(a) sin

6

(b) cos

4

(c) csc

3

Solution: We have

(a) sin

6

= sin

6 = 1

2 (b) cos

4

= cos

4 =

2

2

(c) csc

3= csc

3 = 2

3

3

Fundamental Identities

Proof: The reciprocal identities follow immediately from the definition. We now prove thePythagorean identities. By definition, cos t= xand sin t= y,wherexandy are the coordinatesof a point P(x, y) on the unit circle. SinceP(x, y) is on the unit circle, we have x2 +y2 = 1.

Thussin2 t+ cos2 t= 1

Dividing both sides by cos2 t(provided cos t = 0), we get

sin2 t

cos2 t+

cos2 t

cos2 t=

1

cos2 t

sin2 t

cos2 t

2+ 1 =

1

cos2 t

2

tan

2

t+ 1 = sec

2

t

We have used the reciprocal identities sin t/ cos t= tan tand 1/ cos t= sec t.Similarly, dividingboth sides of the first Pythagorean identity by sin2 t (provided sin t= 0) gives us 1 + cot2 t=csc2 t.

EXAMPLE: If cos t = 3

5 and t is in Quadrant IV, find the values of all the trigonometric

functions at t.

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5 and t is in Quadrant IV, find the values of all the trigonometric

functions at t.

Solution: From the Pythagorean identities we have

sin2 t+ cos2 t= 1

sin2 t+

3

5

2= 1

sin2 t= 1 925

=16

25

sin t= 45

Since this point is in Quadrant IV, sin t is negative, so sin t =45

. Now that we know both

sin t and cos t, we can find the values of the other trigonometric functions using the reciprocalidentities:

sin t=

4

5

cos t=3

5

tan t= sin t

cos t

= 4

5

3

5

=

4

3

csc t= 1

sin t= 5

4 sec t=

1

cos t=

5

3 cot t=

1

tan t= 3

4


and t is in Quadrant II, find the values of all the trigonometric

functions at t.

Solution: From the Pythagorean identities we have

sin2 t+ cos2 t= 1

sin2 t+

5

13

2= 1

sin2 t= 1 25169

=144

169

sin t= 1213

Since this point is in Quadrant II, sin t is positive, so sin t = 12

13. Now that we know both

sin t and cos t, we can find the values of the other trigonometric functions using the reciprocal

identities:

sin t=12

13 cos t= 5

13 tan t=

sin t

cos t=

12

13

513

= 125

csc t= 1

sin t=

13

12 sec t=

1

cos t= 13

5 cot t=

1

tan t= 5

12

EXAMPLE: Write tan t in terms of cos t, where t is in Quadrant III.

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EXAMPLE: Write tan t in terms of cos t, where t is in Quadrant III.

Solution: Since tan t= sin t/ cos t,we need to write sin t in terms of cos t. By the Pythagoreanidentities we have

sin2 t+ cos2 t= 1

sin2 t= 1 cos2 t

sin t=

1

cos2 t

Since sin t is negative in Quadrant III, the negative sign applies here. Thus

tan t= sin t

cos t=

1 cos2 t

cos t

EXAMPLE: Write tan t in terms of cos t, where t is in Quadrant I.


sin2 t+ cos2 t= 1

sin2 t= 1

cos2 t

sin t=

1 cos2 tSince sin t is positive in Quadrant I, the positive sign applies here. Thus

tan t= sin t

cos t=

1 cos2 t

cos t

EXAMPLE: Write cos t in terms of tan t, where t is in Quadrant II.


sin2 t+ cos2 t= 1

sin2 t= 1 cos2 t

so tan2 t= sin2 t

cos2 t=

1 cos2 tcos2 t

.Multiplying both sides by cos2 t, we get

cos2 t tan2 t= 1 cos2 tcos2 t tan2 t+ cos2 t= 1

cos2 t(tan2 t+ 1) = 1

cos2 t= 1tan2 t+ 1

cos t= 1tan2 t+ 1

Since cos tis negative in Quadrant II, the negative sign applies here. Thus

cos t= 1tan2 t+ 1

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Trigonometric Functions of Real Numbers