Section 5.2 Trigonometric Functions of Real Numbers

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1 Section. 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. a) t = π b) t = π 1

EXAMPLE: Use the Table below to find the six trigonometric functions of each given real number t. a) t = π b) t = π a) From the Table, we see that the terminal point determined by t = π/ is P1/, /).

So sin π = 1 cos π = 0 csc π = 1 1 = 1 cot π = 0 1 = 0 But tanπ/ and secπ/ are undefined because x = 0 appears in the denominator in each of their definitions.

2 EXAMPLE: Use the Table below to find the six trigonometric functions of each given real number t. a) t = π b) t = π a) From the Table, we see that the terminal point determined by t = π/ is P1/, /). Since the coordinates are x = 1/ and y = /, we have sin π = csc π = cos π = 1 tan π = / 1/ = sec π = cot π = 1/ / = b) The terminal point determined by π/ is P0,1). So sin π = 1 cos π = 0 csc π = 1 1 = 1 cot π = 0 1 = 0 But tanπ/ and secπ/ are undefined because x = 0 appears in the denominator in each of their definitions. EXAMPLE: Find the six trigonometric functions of each given real number t = π. From the Table above, we see that the terminal point determined by t = π/ is P /, /). Since the coordinates are x = / and y = /, we have sin π = cos π = tan π = / / = 1 csc π = sec π = cot π = / / = 1

3 Values of the Trigonometric Functions EXAMPLE: a) cos π > 0, because the terminal point of t = π is in Quadrant I. b) tan > 0, because the terminal point of t = is in Quadrant III. c) If cost < 0 and sint > 0, then the terminal point of t must be in Quadrant II. EXAMPLE: Determine the sign of each function. a) cos 7π a) Positive b) tan1 b) Positive EXAMPLE: Find each value. a) cos π b) tan π ) c) sin 19π

c) If cost < 0 and sint > 0, then the terminal point of t must be in Quadrant II.

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

4 EXAMPLE: Find each value. a) cos π b) tan π ) c) sin 19π a) Since π π π = = π π = π π the reference number for π/ is π/ see Figure a) below) and the terminal point of π/ is in Quadrant II. Thus cosπ/) is negative and b) The reference number for π/ is π/ see Figure b) below). Since the terminal point of π/ is in Quadrant IV, tan π/) is negative. Thus c) Since 19π 0π π = = 0π π = π π the reference number for 19π/ is π/ see Figure c) below) and the terminal point of 19π/ is in Quadrant II. Thus sin19π/) is positive and EXAMPLE: Find each value. a) sin π b) tan π ) c) cos 1π

is in Quadrant II. Thus cosπ/) is negative and b) The reference number for π/ is π/ see Figure b) below).

5 EXAMPLE: Find each value. a) sin π b) tan π ) c) cos 1π a) Since π π π = = π π = π π the reference number for π/ is π/ and the terminal point of π/ is in Quadrant II. Thus sinπ/) is positive and sin π = sin π = b) Since π +π = π = π π = π π the reference number for π/ is π/ and the terminal point of π/ is in Quadrant II. Thus tan π/) is negative and tan π ) π = tan = ) c) Since 1π 1π π = = 1π π = π π the reference number for 1π/ is π/ and the terminal point of 1π/ is in Quadrant II. Thus cos1π/) is negative and cos 1π = cos π = 1 EXAMPLE: Evaluate a) sin π b) cos 7π c) tan 11π d) sec 17π e) csc 17π f) cot 11π

Thus sinπ/) is positive and sin π = sin π = b) Since π +π = π = π π = π π the reference number for π/ is π/ and the terminal point of π/ is in Quadrant

EXAMPLE: Evaluate a) sin π b) cos 7π c) tan 11π d) sec 17π e) csc 17π f) cot 11π a) The reference number for π/ is π/. Since the terminal point of π/ is in Quadrant I, sinπ/) is positive.

6 EXAMPLE: Evaluate a) sin π b) cos 7π c) tan 11π d) sec 17π e) csc 17π f) cot 11π a) The reference number for π/ is π/. Since the terminal point of π/ is in Quadrant I, sinπ/) is positive. Thus sin π = sin π = b) Since 7π = π+π = π + π = π + π, the reference number for 7π/ is π/ and the terminal point of 7π/ is in Quadrant III. Thus cos7π/) is negative and cos 7π = cos π = c) Since 11π = 1π π = 1π π = π π, the reference number for 11π/ is π/ and the terminal point of 11π/ is in Quadrant II. Thus tan11π/) is negative and tan 11π = tan π = 1 d) Since 17π = 18π π = 18π π = π π, the reference number for 17π/ is π/ and the terminal point of 17π/ is in Quadrant IV. Thus sec17π/) is positive and sec 17π = sec π = e) Since 17π = 1π+π = 1π + π = 8π + π, the reference number for 17π/ is π/ and the terminal point of 17π/ is in Quadrant I II). Thus csc17π/) is positive and csc 17π = csc π = 1 f) Since 11π = 10π+π = 10π + π = 0π + π, the reference number for 11π/ is π/ and the terminal point of 11π/ is in Quadrant I. Thus cot11π/) is positive and cot 11π = cot π = EXAMPLE: Use the even-odd properties of the trigonometric functions to determine each value. a) sin π ) b) cos π ) c) csc π )

Thus cos7π/) is negative and cos 7π = cos π = c) Since 11π = 1π π = 1π π = π π, the reference number for 11π/ is π/ and the terminal point of 11π/ is in Quadrant II.

7 EXAMPLE: Use the even-odd properties of the trigonometric functions to determine each value. a) sin π ) b) cos π ) c) csc π ) We have a) sin π ) = sin π = 1 c) csc π ) = csc π = b) cos π ) = cos π = Fundamental Identities Proof: The reciprocal identities follow immediately from the definition. We now prove the Pythagorean identities. Bydefinition, cost = xand sint = y,wherexandy arethecoordinates of a point Px,y) on the unit circle. Since Px,y) is on the unit circle, we have x +y = 1. Thus sin t+cos t = 1 Dividing both sides by cos t provided cost 0), we get sin t cos t + cos t cos t = 1 cos t ) sin t 1 +1 = cos t cos t tan t+1 = sec t We have used the reciprocal identities sint/cost = tant and 1/cost = sect. Similarly, dividing both sides of the first Pythagorean identity by sin t provided sint 0) gives us 1+cot t = csc t. ) EXAMPLE: If cost = functions at t. and t is in Quadrant IV, find the values of all the trigonometric 7

definition. We now prove the Pythagorean identities. Bydefinition, cost = xand sint = y,wherexandy arethecoordinates of a point Px,y) on the unit circle.

8 EXAMPLE: If cost = functions at t. and t is in Quadrant IV, find the values of all the trigonometric From the Pythagorean identities we have sin t+cos t = 1 ) sin t+ = 1 sin t = 1 9 = 1 sint = ± Since this point is in Quadrant IV, sint is negative, so sint =. Now that we know both sint and cost, we can find the values of the other trigonometric functions using the reciprocal identities: sint = csct = 1 sint = cost = sect = 1 cost = tant = sint cost = cott = 1 tant = = EXAMPLE: If cost = 1 functions at t. and t is in Quadrant II, find the values of all the trigonometric From the Pythagorean identities we have sin t+ sin t+cos t = 1 ) = 1 1 sin t = 1 19 = 1 19 sint = ± 1 1 Since this point is in Quadrant II, sint is positive, so sint = 1. Now that we know both 1 sint and cost, we can find the values of the other trigonometric functions using the reciprocal identities: sint = 1 1 csct = 1 sint = 1 1 cost = 1 sect = 1 cost = 1 tant = sint 1 cost = 1 EXAMPLE: Write tant in terms of cost, where t is in Quadrant III. 8 1 cott = 1 tant = 1 = 1

9 EXAMPLE: Write tant in terms of cost, where t is in Quadrant III. Since tant = sint/cost, we need to write sint in terms of cost. By the Pythagorean identities we have sin t+cos t = 1 sin t = 1 cos t sint = ± 1 cos t Since sint is negative in Quadrant III, the negative sign applies here. Thus tant = sint cost = 1 cos t cost EXAMPLE: Write tant in terms of cost, where t is in Quadrant I. Since tant = sint/cost, we need to write sint in terms of cost. By the Pythagorean identities we have sin t+cos t = 1 sin t = 1 cos t sint = ± 1 cos t Since sint is positive in Quadrant I, the positive sign applies here. Thus tant = sint cost = 1 cos t cost EXAMPLE: Write cost in terms of tant, where t is in Quadrant II. Since tant = sint/cost, we need to write sint in terms of cost. By the Pythagorean identities we have sin t+cos t = 1 sin t = 1 cos t so tan t = sin t cos t = 1 cos t. Multiplying both sides by cos t, we get cos t cos ttan t = 1 cos t cos ttan t+cos t = 1 cos ttan t+1) = 1 cos t = 1 tan t+1 1 cost = ± tan t+1 Since cost is negative in Quadrant II, the negative sign applies here. Thus 1 cost = tan t+1 9

Thus tant = sint cost = 1 cos t cost EXAMPLE: Write tant in terms of cost, where t is in Quadrant I. Since tant = sint/cost, we need to write sint in terms of cost.

4.3 & 4.8 Right Triangle Trigonometry. Anatomy of Right Triangles

4.3 & 4.8 Right Triangle Trigonometry Anatomy of Right Triangles The right triangle shown at the right uses lower case a, b and c for its sides with c being the hypotenuse. The sides a and b are referred