# Trigonometric Functions

- Pages:
**9** - Word count:
**2064** - Category: Mathematics

A limited time offer! Get a custom sample essay written according to your requirements urgent 3h delivery guaranteed

Order NowEXAMPLE: Use the Table below to ο¬nd the six trigonometric functions of each given real number t. Ο Ο (a) t = (b) t = 3 2

EXAMPLE: Use the Table below to ο¬nd 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 β t = β is P (1/2, 3/2). Since the coordinates are x = 1/2 and Ο/3 y = 3/2, we have β β Ο 3 3/2 β Ο 1 Ο sin = cos = tan = = 3 3 2 3 2 3 1/2 β β Ο 3 2 3 Ο Ο 1/2 csc = = sec = 2 cot = β 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 undeο¬ned because x = 0 appears in the denominator in each of their deο¬nitions. Ο . 4 Solution: β From the Table above, we see that β terminal point determined by t = Ο/4 is the β β P ( 2/2, 2/2). Since the coordinates are x = 2/2 and y = 2/2, we have β β β Ο 2 2 2/2 Ο Ο sin = =1 cos = tan = β 4 2 4 2 4 2/2 β Ο β Ο Ο β 2/2 csc = 2 sec = 2 cot = β =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 β 3 3 19Ο 4

(c) sin

EXAMPLE: Find each value. Ο 19Ο 2Ο (b) tan β (c) sin (a) cos 3 3 4 Solution: (a) Since 2Ο 3Ο β Ο 3Ο Ο Ο = = β =Οβ 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 βΟ/3 is Ο/3 (see Figure (b) below). Since the terminal point of βΟ/3 is in Quadrant IV, tan(βΟ/3) is negative. Thus

(c) Since

19Ο 20Ο β Ο 20Ο Ο Ο = = β = 5Ο β 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 β 3 3

(c) cos

14Ο 3

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

(c) cos

14Ο 3

2Ο 3Ο β Ο 3Ο Ο Ο = = β =Οβ 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 β 2Ο 3 Ο sin = sin = 3 3 2 (b) Since 4Ο 3Ο + Ο 3Ο Ο Ο =β =β β = βΟ β 3 3 3 3 3 the reference number for β4Ο/3 is Ο/3 and the terminal point of β4Ο/3 is in Quadrant II. Thus tan (β4Ο/3) is negative and β tan β (c) Since 4Ο 3 = β

tan Ο 3 β =β 3

14Ο 15Ο β Ο 15Ο Ο Ο = = β = 5Ο β 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 = β cos = β 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 β 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 β Ο 7Ο 3 = β cos = β cos 6 6 2 (c) Since 11Ο = 12ΟβΟ = 12Ο β Ο = 3Ο β 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 = β tan = β1 4 4

(d) Since 17Ο = 18ΟβΟ = 18Ο β Ο = 6Ο β 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 β 121Ο = cot = 3 6 6

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

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

Fundamental Identities

Proof: The reciprocal identities follow immediately from the deο¬nition. We now prove the Pythagorean identities. By deο¬nition, 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 ο¬rst Pythagorean identity by sin2 t (provided sin t = 0) gives us 1 + cot2 t = csc2 t. 3 and t is in Quadrant IV, ο¬nd the values of all the trigonometric EXAMPLE: If cos t = 5 functions at t. 7

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

=1 9 16 = 25 25

sin2 t = 1 β sin t = Β± 4 5

4 Since this point is in Quadrant IV, sin t is negative, so sin t = β . Now that we know both 5 sin t and cos t, we can ο¬nd the values of the other trigonometric functions using the reciprocal identities: sin t = β csc t = 4 5 cos t = sec t = 3 5 1 5 = cos t 3 tan t = cot t = β4 4 sin t = 35 = β cos t 3 5 1 3 =β tan t 4

1 5 =β sin t 4

5 and t is in Quadrant II, ο¬nd the values of all the trigonometric EXAMPLE: If cos t = β 13 functions at t. Solution: From the Pythagorean identities we have sin2 t + cos2 t = 1 sin2 t + β 5 13 2Β =1

sin2 t = 1 β

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 ο¬nd the values of the other trigonometric functions using the reciprocal identities: sin t = csc t = 12 13 1 13 = sin t 12 cos t = β sec t = 5 13 12 12 sin t = 135 = β cos t 5 β 13 1 5 cot t = =β tan t 12

tan t =

1 13 =β 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 β cos2 t β sin t = Β± 1 β cos2 t

Since sin t is negative in Quadrant III, the negative sign applies here. Thus β 1 β cos2 t sin t =β 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 β cos2 t β sin t = Β± 1 β cos2 t

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

so tan2 t =

1 β cos2 t sin2 t = . Multiplying both sides by cos2 t, we get cos2 t cos2 t cos2 t tan2 t = 1 β 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 = β β 2 tan t + 1 9

cos t = Β± β