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Trigonometric Functions

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EXAMPLE: Use the Table below to find the six trigonometric functions of each given real number t. Ο€ Ο€ (a) t = (b) t = 3 2

EXAMPLE: Use the Table below to find the six trigonometric functions of each given real number t. Ο€ Ο€ (a) t = (b) t = 3 2 Solution: (a) From the Table, we see that the terminal point determined by √ t = √ is P (1/2, 3/2). Since the coordinates are x = 1/2 and Ο€/3 y = 3/2, we have √ √ Ο€ 3 3/2 √ Ο€ 1 Ο€ sin = cos = tan = = 3 3 2 3 2 3 1/2 √ √ Ο€ 3 2 3 Ο€ Ο€ 1/2 csc = = sec = 2 cot = √ 3 3 3 3 3 3/2 (b) The terminal point determined by Ο€/2 is P (0, 1). So Ο€ Ο€ 1 Ο€ 0 Ο€ cos = 0 csc = = 1 cot = = 0 sin = 1 2 2 2 1 2 1 But tan Ο€/2 and sec Ο€/2 are undefined because x = 0 appears in the denominator in each of their definitions. Ο€ . 4 Solution: √ From the Table above, we see that √ terminal point determined by t = Ο€/4 is the √ √ P ( 2/2, 2/2). Since the coordinates are x = 2/2 and y = 2/2, we have √ √ √ Ο€ 2 2 2/2 Ο€ Ο€ sin = =1 cos = tan = √ 4 2 4 2 4 2/2 √ Ο€ √ Ο€ Ο€ √ 2/2 csc = 2 sec = 2 cot = √ =1 4 4 4 2/2 EXAMPLE: Find the six trigonometric functions of each given real number t =

Values of the Trigonometric Functions

EXAMPLE: Ο€ Ο€ (a) cos > 0, because the terminal point of t = is in Quadrant I. 3 3 (b) tan 4 > 0, because the terminal point of t = 4 is in Quadrant III. (c) If cos t < 0 and sin t > 0, then the terminal point of t must be in Quadrant II. EXAMPLE: Determine the sign of each function. 7Ο€ (b) tan 1 (a) cos 4 Solution: (a) Positive (b) Positive EXAMPLE: Find each value. 2Ο€ Ο€ (a) cos (b) tan βˆ’ 3 3 19Ο€ 4

(c) sin

EXAMPLE: Find each value. Ο€ 19Ο€ 2Ο€ (b) tan βˆ’ (c) sin (a) cos 3 3 4 Solution: (a) Since 2Ο€ 3Ο€ βˆ’ Ο€ 3Ο€ Ο€ Ο€ = = βˆ’ =Ο€βˆ’ 3 3 3 3 3 the reference number for 2Ο€/3 is Ο€/3 (see Figure (a) below) and the terminal point of 2Ο€/3 is in Quadrant II. Thus cos(2Ο€/3) is negative and

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

(c) Since

19Ο€ 20Ο€ βˆ’ Ο€ 20Ο€ Ο€ Ο€ = = βˆ’ = 5Ο€ βˆ’ 4 4 4 4 4 the reference number for 19Ο€/4 is Ο€/4 (see Figure (c) below) and the terminal point of 19Ο€/4 is in Quadrant II. Thus sin(19Ο€/4) is positive and

EXAMPLE: Find each value. 2Ο€ 4Ο€ (a) sin (b) tan βˆ’ 3 3

(c) cos

14Ο€ 3

EXAMPLE: Find each value. 4Ο€ 2Ο€ (b) tan βˆ’ (a) sin 3 3 Solution: (a) Since

(c) cos

14Ο€ 3

2Ο€ 3Ο€ βˆ’ Ο€ 3Ο€ Ο€ Ο€ = = βˆ’ =Ο€βˆ’ 3 3 3 3 3 the reference number for 2Ο€/3 is Ο€/3 and the terminal point of 2Ο€/3 is in Quadrant II. Thus sin(2Ο€/3) is positive and √ 2Ο€ 3 Ο€ sin = sin = 3 3 2 (b) Since 4Ο€ 3Ο€ + Ο€ 3Ο€ Ο€ Ο€ =βˆ’ =βˆ’ βˆ’ = βˆ’Ο€ βˆ’ 3 3 3 3 3 the reference number for βˆ’4Ο€/3 is Ο€/3 and the terminal point of βˆ’4Ο€/3 is in Quadrant II. Thus tan (βˆ’4Ο€/3) is negative and βˆ’ tan βˆ’ (c) Since 4Ο€ 3 = βˆ’
tan Ο€ 3 √ =βˆ’ 3

14Ο€ 15Ο€ βˆ’ Ο€ 15Ο€ Ο€ Ο€ = = βˆ’ = 5Ο€ βˆ’ 3 3 3 3 3 the reference number for 14Ο€/3 is Ο€/3 and the terminal point of 14Ο€/3 is in Quadrant II. Thus cos(14Ο€/4) is negative and 14Ο€ Ο€ 1 cos = βˆ’ cos = βˆ’ 3 3 2 EXAMPLE: Evaluate 7Ο€ Ο€ (b) cos (a) sin 3 6 11Ο€ 4 17Ο€ 3 17Ο€ 2 121Ο€ 6

(c) tan

(d) sec

(e) csc

(f) cot

EXAMPLE: Evaluate 7Ο€ 11Ο€ 17Ο€ 17Ο€ 121Ο€ Ο€ (b) cos (c) tan (d) sec (e) csc (f) cot (a) sin 3 6 4 3 2 6 Solution: (a) The reference number for Ο€/3 is Ο€/3. Since the terminal point of Ο€/3 is in Quadrant I, sin(Ο€/3) is positive. Thus √ Ο€ 3 Ο€ sin = sin = 3 3 2 (b) Since 7Ο€ = 6Ο€+Ο€ = 6Ο€ + Ο€ = Ο€ + Ο€ , the reference number for 7Ο€/6 is Ο€/6 and the terminal 6 6 6 6 6 point of 7Ο€/6 is in Quadrant III. Thus cos(7Ο€/6) is negative and √ Ο€ 7Ο€ 3 = βˆ’ cos = βˆ’ cos 6 6 2 (c) Since 11Ο€ = 12Ο€βˆ’Ο€ = 12Ο€ βˆ’ Ο€ = 3Ο€ βˆ’ Ο€ , the reference number for 11Ο€/4 is Ο€/4 and the 4 4 4 4 4 terminal point of 11Ο€/6 is in Quadrant II. Thus tan(11Ο€/4) is negative and tan Ο€ 11Ο€ = βˆ’ tan = βˆ’1 4 4

(d) Since 17Ο€ = 18Ο€βˆ’Ο€ = 18Ο€ βˆ’ Ο€ = 6Ο€ βˆ’ Ο€ , the reference number for 17Ο€/3 is Ο€/3 and the 3 3 3 3 3 terminal point of 17Ο€/3 is in Quadrant IV. Thus sec(17Ο€/3) is positive and sec 17Ο€ Ο€ = sec = 2 3 3

(e) Since 17Ο€ = 16Ο€+Ο€ = 16Ο€ + Ο€ = 8Ο€ + Ο€ , the reference number for 17Ο€/2 is Ο€/2 and the 2 2 2 2 2 terminal point of 17Ο€/2 is in Quadrant I (II). Thus csc(17Ο€/2) is positive and csc Ο€ 17Ο€ = csc = 1 2 2

(f) Since 121Ο€ = 120Ο€+Ο€ = 120Ο€ + Ο€ = 20Ο€ + Ο€ , the reference number for 121Ο€/6 is Ο€/6 and the 6 6 6 6 6 terminal point of 121Ο€/6 is in Quadrant I. Thus cot(121Ο€/6) is positive and cot Ο€ √ 121Ο€ = cot = 3 6 6

EXAMPLE: Use the even-odd properties of the trigonometric functions to determine each value. Ο€ Ο€ Ο€ (b) cos βˆ’ (c) csc βˆ’ (a) sin βˆ’ 6 4 3 6

EXAMPLE: Use the even-odd properties of the trigonometric functions to determine each value. Ο€ Ο€ Ο€ (a) sin βˆ’ (b) cos βˆ’ (c) csc βˆ’ 6 4 3 Solution: We have √ Ο€ 1 Ο€ Ο€ Ο€ 2 = βˆ’ sin = βˆ’ (b) cos βˆ’ = cos = (a) sin βˆ’ 6 6 2 4 4 2 √ Ο€ 2 3 Ο€ (c) csc βˆ’ = βˆ’ csc = βˆ’ 3 3 3

Fundamental Identities

Proof: The reciprocal identities follow immediately from the definition. We now prove the Pythagorean identities. By definition, cos t = x and sin t = y, where x and y are the coordinates of a point P (x, y) on the unit circle. Since P (x, y) is on the unit circle, we have x2 + y 2 = 1. Thus sin2 t + cos2 t = 1 Dividing both sides by cos2 t (provided cos t = 0), we get sin2 t cos2 t 1 + = 2t 2t cos cos cos2 t sin2 t cos2 t 2

+1=

1 cos2 t

tan2 t + 1 = sec2 t We have used the reciprocal identities sin t/ cos t = tan t and 1/ cos t = sec t. Similarly, dividing both sides of the first Pythagorean identity by sin2 t (provided sin t = 0) gives us 1 + cot2 t = csc2 t. 3 and t is in Quadrant IV, find the values of all the trigonometric EXAMPLE: If cos t = 5 functions at t. 7

3 and t is in Quadrant IV, find the values of all the trigonometric EXAMPLE: If cos t = 5 functions at t. Solution: From the Pythagorean identities we have sin2 t + cos2 t = 1 sin2 t + 3 5 2

=1 9 16 = 25 25

sin2 t = 1 βˆ’ sin t = Β± 4 5

4 Since this point is in Quadrant IV, sin t is negative, so sin t = βˆ’ . Now that we know both 5 sin t and cos t, we can find the values of the other trigonometric functions using the reciprocal identities: sin t = βˆ’ csc t = 4 5 cos t = sec t = 3 5 1 5 = cos t 3 tan t = cot t = βˆ’4 4 sin t = 35 = βˆ’ cos t 3 5 1 3 =βˆ’ tan t 4

1 5 =βˆ’ sin t 4

5 and t is in Quadrant II, find the values of all the trigonometric EXAMPLE: If cos t = βˆ’ 13 functions at t. Solution: From the Pythagorean identities we have sin2 t + cos2 t = 1 sin2 t + βˆ’ 5 13 2Β =1

sin2 t = 1 βˆ’

25 144 = 169 169 12 sin t = Β± 13

12 Since this point is in Quadrant II, sin t is positive, so sin t = . Now that we know both 13 sin t and cos t, we can find the values of the other trigonometric functions using the reciprocal identities: sin t = csc t = 12 13 1 13 = sin t 12 cos t = βˆ’ sec t = 5 13 12 12 sin t = 135 = βˆ’ cos t 5 βˆ’ 13 1 5 cot t = =βˆ’ tan t 12

tan t =

1 13 =βˆ’ cos t 5

EXAMPLE: Write tan t in terms of cos t, where t is in Quadrant III. 8

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 Pythagorean identities we have sin2 t + cos2 t = 1 sin2 t = 1 βˆ’ cos2 t √ sin t = Β± 1 βˆ’ cos2 t

Since sin t is negative in Quadrant III, the negative sign applies here. Thus √ 1 βˆ’ cos2 t sin t =βˆ’ tan t = cos t cos t EXAMPLE: Write tan t in terms of cos t, where t is in Quadrant I. Solution: Since tan t = sin t/ cos t, we need to write sin t in terms of cos t. By the Pythagorean identities we have sin2 t + cos2 t = 1 sin2 t = 1 βˆ’ cos2 t √ sin t = Β± 1 βˆ’ cos2 t

Since sin t is positive in Quadrant I, the positive sign applies here. Thus √ 1 βˆ’ cos2 t sin t = tan t = cos t cos t EXAMPLE: Write cos t in terms of tan t, where t is in Quadrant II. Solution: Since tan t = sin t/ cos t, we need to write sin t in terms of cos t. By the Pythagorean identities we have sin2 t + cos2 t = 1 sin2 t = 1 βˆ’ cos2 t

so tan2 t =

1 βˆ’ cos2 t sin2 t = . Multiplying both sides by cos2 t, we get cos2 t cos2 t cos2 t tan2 t = 1 βˆ’ cos2 t cos2 t tan2 t + cos2 t = 1 cos2 t(tan2 t + 1) = 1 cos2 t = 1 tan t + 1 2

tan2 t + 1 Since cos t is negative in Quadrant II, the negative sign applies here. Thus 1 cos t = βˆ’ √ 2 tan t + 1 9

cos t = ± √

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