Trigonometric Functions

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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 real number t. π π (a) t = (b) t = 3 2

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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 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 =

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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

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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

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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

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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...
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