Trigonometric Functions

Section five. 2 Trigonometric Functions of Real Quantities

The Trigonometric Functions

MODEL: Use the Table below to find the six trigonometric functions of each and every given true number t. π π (a) t = (b) t sama dengan 3 two

1

EXAMPLE: Use the Stand below to find the six trigonometric functions of each and every given true number t. π π (a) to = (b) t = 3 a couple of Solution: (a) From the Stand, we see the fact that terminal stage determined by √ t = √ can be P (1/2, 3/2). Considering that the coordinates happen to be x sama dengan 1/2 and π/3 sumado a = 3/2, we have √ √ π 3 3/2 √ π 1 π sin sama dengan cos = tan = = several 3 2 3 a couple of 3 1/2 √ √ π a few 2 3 π π 1/2 csc = sama dengan sec sama dengan 2 crib = √ 3 three or more 3 three or more 3 3/2 (b) The terminal level determined by π/2 is L (0, 1). So π π 1 π zero π cos = zero csc = = one particular cot sama dengan = 0 sin sama dengan 1 2 2 two 1 2 1 But tan π/2 and securities and exchange commission's π/2 will be undefined because x = 0 shows up in the denominator in every of their definitions. π. four Solution: √ From the Stand above, we see that √ terminal stage determined by t = π/4 is the √ √ G ( 2/2, 2/2). Considering that the coordinates will be x = 2/2 and y = 2/2, we now have √ √ √ π 2 2 2/2 π π trouble = =1 cos sama dengan tan = √ 4 2 5 2 four 2/2 √ π √ π π √ 2/2 csc = 2 securities and exchange commission's = a couple of cot sama dengan √ =1 4 4 4 2/2 EXAMPLE: Discover the half a dozen trigonometric features of each offered real number t =

2

Ideals of the Trigonometric Functions

CASE IN POINT: π π (a) cos > 0, because the airport terminal point of t sama dengan is in Sector I. three or more 3 (b) tan 4 > zero, because the terminal point of t sama dengan 4 is at Quadrant III. (c) If perhaps cos capital t < 0 and trouble t > 0, then this terminal stage of t must be in Quadrant 2. EXAMPLE: Identify the signal of each function. 7π (b) tan you (a) cos 4 Option: (a) Positive (b) Confident EXAMPLE: Find each worth. 2π π (a) cos (b) bronze − 3 3 19π 4

(c) sin

a few

EXAMPLE: Locate each value. π 19π 2π (b) tan − (c) desprovisto (a) cos 3 three or more 4 Solution: (a) Since 2π 3π − π 3π π π sama dengan = − =π− a few 3 three or more 3 three or more the reference number for 2π/3 is π/3 (see Number (a) below) and the terminal point of 2π/3 is in Quadrant 2. Thus cos(2π/3) is negative and

(b) The client name for −π/3 is π/3 (see Figure (b) below). Since the fatal point of −π/3 is within Quadrant IV, tan(−π/3) is usually negative. Thus

(c) Seeing that

19π 20π − π 20π π π = = − = 5π − 5 4 5 4 4 the client name for 19π/4 is π/4 (see Physique (c) below) and the port point of 19π/4 is within Quadrant 2. Thus sin(19π/4) is positive and

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

(c) cos

14π a few

4

MODEL: Find every single value. 4π 2π (b) tan − (a) trouble 3 3 Solution: (a) Since

(c) cos

14π 3

2π 3π − π 3π π π = sama dengan − =π− 3 several 3 three or more 3 the reference number for 2π/3 can be π/3 and the terminal level of 2π/3 is in Quadrant II. Thus sin(2π/3) can be positive and √ 2π 3 π sin sama dengan sin = 3 3 2 (b) Since 4π 3π & π 3π π π =− =− − sama dengan −π − 3 several 3 3 3 the reference number pertaining to −4π/3 is definitely π/3 as well as the terminal level of −4π/3 is in Particular II. Thus tan (−4π/3) is unfavorable and − tan − (c) As 4π 3 = − tan π 3 √ =− several

14π 15π − π 15π π π = = − = 5π − several 3 3 3 a few the client name for 14π/3 is π/3 and the port point of 14π/3 is Quadrant II. Thus cos(14π/4) is negative and 14π π you cos sama dengan − cos = − 3 three or more 2 MODEL: Evaluate 7π π (b) cos (a) sin several 6 11π 4 17π 3 17π 2 121π 6

(c) tan

(d) sec

(e) csc

(f) cot

5

EXAMPLE: Assess 7π 11π 17π 17π 121π π (b) cos (c) bronze (d) sec (e) csc (f) cot (a) desprovisto 3 6 4 a few 2 6 Solution: (a) The reference number for π/3 is π/3. Since the port point of π/3 is within Quadrant I actually, sin(π/3) is definitely positive. As a result √ π 3 π sin = sin = 3 three or more 2 (b) Since 7π = 6π+π = 6π + π = π + π, the reference number for 7π/6 is π/6 and the fatal 6 6th 6 six 6 stage of 7π/6 is in Sector III. As a result cos(7π/6) can be negative and √ π 7π several = − cos sama dengan − cos 6 six 2 (c) Since 11π = 12π−π = 12π − π = 3π − π, the client name for 11π/4 is π/4 and the some 4 5 4 four terminal level of 11π/6 is in Quadrant II. As a result tan(11π/4) can be negative and tan π 11π = − suntan = −1 4 some

(d) Since 17π = 18π−π sama dengan 18π − π sama dengan 6π − π, the reference number intended for 17π/3 is usually π/3 and the 3 several 3 a few 3 airport terminal point of 17π/3 is in Quadrant IV. Thus sec(17π/3) is...

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