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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. π π (a) t = (b) t = 3 2

EXAMPLE: Use the Table below to ﬁnd the six trigonometric functions of each given real number t. π π (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 π/3 y = 3/2, we have √ √ π 3 3/2 √ π 1 π sin = cos = tan = = 3 3 2 3 2 3 1/2 √ √ π 3 2 3 π π 1/2 csc = = sec = 2 cot = √ 3 3 3 3 3 3/2 (b) The terminal point determined by π/2 is P (0, 1). So π π 1 π 0 π cos = 0 csc = = 1 cot = = 0 sin = 1 2 2 2 1 2 1 But tan π/2 and sec π/2 are undeﬁned because x = 0 appears in the denominator in each of their deﬁnitions. π . 4 Solution: √ From the Table above, we see that √ terminal point determined by t = π/4 is the √ √ P ( 2/2, 2/2). Since the coordinates are x = 2/2 and y = 2/2, we have √ √ √ π 2 2 2/2 π π sin = =1 cos = tan = √ 4 2 4 2 4 2/2 √ π √ π π √ 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: π π (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. 7π (b) tan 1 (a) cos 4 Solution: (a) Positive (b) Positive EXAMPLE: Find each value. 2π π (a) cos (b) tan − 3 3 19π 4

(c) sin

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

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

(c) Since

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

EXAMPLE: Find each value. 2π 4π (a) sin (b) tan − 3 3

(c) cos

14π 3

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

(c) cos

14π 3

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

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

(c) tan

(d) sec

(e) csc

(f) cot

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

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

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

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

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

EXAMPLE: Use the even-odd properties of the trigonometric functions to determine each value. π π π (a) sin − (b) cos − (c) csc − 6 4 3 Solution: We have √ π 1 π π π 2 = − sin = − (b) cos − = cos = (a) sin − 6 6 2 4 4 2 √ π 2 3 π (c) csc − = − csc = − 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 − sin t = ± 4 5

4 Since this point is in Quadrant IV, sin t is negative, so sin t = − . 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 = − csc t = 4 5 cos t = sec t = 3 5 1 5 = cos t 3 tan t = cot t = −4 4 sin t = 35 = − cos t 3 5 1 3 =− tan t 4

1 5 =− sin t 4

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

sin2 t = 1 −

25 144 = 169 169 12 sin t = ± 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 = − sec t = 5 13 12 12 sin t = 135 = − cos t 5 − 13 1 5 cot t = =− tan t 12

tan t =

1 13 =− 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 − cos2 t √ sin t = ± 1 − cos2 t

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

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

so tan2 t =

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

cos t = ± √

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