Sect 7.4 Trigonometric Functions of Any Angles

Transcription

1 Sect 7.4 Trigonometric Functions of Any Angles Objective #: Extending the definition to find the trigonometric function of any angle. Before we can extend the definition our trigonometric functions, we need to define an angle in standard position. To get an angle in standard position, we superimpose the angle on a rectangular coordinate system in such a way that the vertex lies on the origin and one of the sides (the initial side) coincides with the positive x-axis. Standard Position An angle θ is said to be in standard position if the vertex is at the origin and the initial side of θ coincides with the positive x-axis. The terminal side is the side of θ that passes through the point (a, b) and r is the distance between the origin and (a, b). (a, b) r θ Trigonometric functions Let θ be in standard position. Let (a, b) be any point other than the origin on the terminal side of θ. If r = a +b equals the distance between (a, b) and the origin, then we can define the six trigonometric functions as follows: sin θ = b r cos θ = a r tan θ = b a csc θ = r sec θ = r cot θ = a b a b provided that a and b are not zero. If a = 0, then the terminal side lies on the y-axis and both tan θ and sec θ are undefined. If b = 0, then the terminal side lies on the x-axis and both cot θ and csc θ are undefined. If (a, b) is in the first quadrant, then our definition matches the definition given in section 7.: r = hyp θ a = adj b = opp

2 Also, the fundamental identities still hold true under this new definition: Fundamental Identities tan θ = sinθ csc θ = cosθ sinθ sec θ = cosθ cot θ = cosθ cot θ = sinθ tanθ sin θ + cos θ = tan θ + = sec θ cot θ + = csc θ Find the exact values of the six trigonometric functions of: Ex. a θ = 0 = 0 Ex. b θ = π = 90 Ex. c θ = π = 80 Ex. d θ = π = 70 a) The terminal side of θ intersects the point (, 0), so the radius r =. Thus, sin(0) = b r = 0 = 0 cos(0) = a r = = tan(0) = b a = 0 = 0 csc(0) = r b = 0 sec(0) = r a = = cot(0) = a b = 0 which is undefined which is undefined b) The terminal side of θ intersects the point (0, ), so the radius r =. Thus, sin( π ) = b = r = cos( π ) = a = 0 r = 0 tan( π ) = b a = 0 which is undefined csc( π ) = r b = = sec( π ) = r a = 0 which is undefined cot( π ) = a b = = c) The terminal side of θ intersects the point (, 0), so the radius r =. Thus, sin(π) = b = 0 r = 0 cos(π) = a = r = tan(π) = b a = 0 = 0 csc(π) = r b = which is undefined 0 sec(π) = r a = = cot(π) = a b = 0 which is undefined

3 d) The terminal side of θ intersects the point (0, ), so the radius r =. Thus, sin( π ) = b r = tan( π ) = b a = 0 = cos( π ) = a r which is undefined = 0 = 0 csc( π ) = r b = = sec( π ) = r a = π which is undefined cot( 0 ) = a b = = We can summarize our results: θ θ sin(θ) cos(θ) tan(θ) csc(θ) sec(θ) cot(θ) undefined undefined π 90 0 undefined undefined 0 π undefined undefined π 70 0 undefined undefined 0 Find the exact values of the six trigonometric functions of θ if the terminal side of θ passes though the following point: Ex. ( 5, ) Since a = 5 and b =, then r = ( 5) +() = = 69 =. sin(θ) = b r tan(θ) = b a = 5 = 5 = sec(θ) = r a = 5 = 5 cos(θ) = a = 5 r = 5 csc(θ) = r b = cot(θ) = a b = 5 = 5 Objective #: Trigonometric Functions of Coterminal Angles. Definition: Two angles are coterminal if they have the same initial and terminal sides. Properties of Coterminal Angles ) Two coterminal angles with differ by integer multiplies of π (60 ) ) The trigonometric value of two coterminal angles are equal.

4 4 Find the following: Ex. 4a sin( 450 ) Ex. 4b tan(5π) A coterminal angle to 450 is A coterminal angle to 5π = 70. is 5π π = π π 5π So, sin( 450 ) = sin(70 ) = So, tan(9π) = tan(π) = 0 Ex. 4c cos( 0 ) Ex. 4d sec( 5π 4 ) A coterminal angle to 0 is A coterminal angle to 5π = 0. is 5π + π = π 4 4 So, cos( 0 ) = cos(0 ) = Ex. 5a cos( 5π 6 So, sec( 5π 4 ) = sec( π 4 ) = cos( π 4 ) = ) Ex. 5b csc(7π) Ex. 5c tan( 7π ) a) cos( 5π 5π ) = cos( (π)) = cos( π ) = b) csc(7π) = csc(7π (π)) = csc(π) which is undefined. c) tan( 7π ) = tan( 7π π) = tan( π ) = Objective #: Determine the Signs of the Trigonometric Functions

5 5 Since sin(θ) = b and csc(θ) = r, then the sine and cosecant function are r b positive in the first and second quadrant and negative in the third and fourth quadrant. Since cos(θ) = a and sec(θ) = r, then the cosine and secant r a are positive in the first and fourth quadrant and negative in the second and third quadrant. Finally, tan(θ) = b a and cot(θ) = a means they are positive b when a and b have the same signs (quadrant I and III) and negative when a and b have opposite signs (quadrant II and IV). A phrase to remember this is "All Suckers Take Calculus." All means all the trigonometric functions are positive in quadrant I, Suckers means only the sine (and hence the cosecant) is positive in quadrant II, Take means that only the tangent (and hence the cotangent) is positive in quadrant III, and Calculus means only the cosine (and hence the secant) is positive in quadrant IV. sin(θ) > 0 all are suckers all csc(θ) > 0 positive tan(θ) > 0 cos(θ) > 0 take calculus cot(θ) > 0 sec(θ) > 0 Given the following information, determine the quadrant θ lies in: Ex. 6a sin(θ) < 0, cos(θ) > 0 Ex. 6b tan(θ) > 0, sin(θ) < 0 Ex. 6c csc(θ) > 0, sec(θ) < 0 a) sin(θ) < 0 when θ is quadrant III and IV and cos(θ) > 0 when θ is quadrant I and IV. The overlap occurs in quadrant IV so θ lies in quadrant IV. b) tan(θ) > 0 when θ is quadrant I and III and sin(θ) < 0 when θ is quadrant III and IV. The overlap occurs in quadrant III so θ lies in quadrant III. c) csc(θ) > 0 when θ is quadrant I and II and sec(θ) < 0 when θ is quadrant II and III. The overlap occurs in quadrant II so θ lies in quadrant II.

6 6 Definition A reference angle θ R is the acute angle formed by the terminal side and the x-axis. TrigFunction(θ) = TrigFunction(θ R ) or TrigFunction(θ R ) depending on the function and the quadrant. Quadrant I Quadrant II θ θ = θ R θ R θ R = π θ Quadrant III Quadrant IV θ θ R θ θ R θ R = θ π θ R = π θ Find the exact value of the following: Ex. 7a sin(0 ) Ex. 7b sec( 7π 4 ) Ex. 7c tan(5 ) Ex. 7d cos( 4π ) a) 0 is in Q III, so θ R = 0 80 = 0. Since sine is negative in Q III, then sin(0 ) = sin(0 ) =. b) 7π 4 is Q IV, so θ R = π 7π 4 = π 4. Since secant is positive in Q IV, then sec( 7π 4 ) = sec( π 4 ) =.

7 7 c) 5 is in Q II, so θ R = 80 5 = 45. Since tangent is negative in Q II, then tan(5 ) = tan(45 ) =. d) 4π is in Q III, so θ R = 4π π = π. Since cosine is negative in Q III, then cos( 4π ) = cos( π ) =. Given the following information, find the remaining five trigonometric functions: Ex. 8 cos(θ) = 0.8 and π < θ < π Since π < θ < π, the angle is in Q IV. Since cos(θ R) = adj hyp = 0.8, then the adjacent side is 0.8 and the hypotenuse is. We can then find the opposite side: opp + adj = hyp opp + (0.8) = opp = opp = 0.6 opp = ± 0.6 Thus, sin(θ R ) = opp hyp = 0.6 function is negative, meaning that sin(θ) = 0.6. tan(θ) = sin(θ) cos(θ) = = 0.75 csc(θ) = sec(θ) = cot(θ) = = sin(θ) cos(θ) = Ex. 9 tan(θ) = 0.6 = =.5 tan(θ) = 0.75 = 4 and cos(θ) < 0 = 0.6 and θ is in Q IV, then the sine Since the tangent is positive and the cosine is negative, the angle is in Q III. Since tan(θ R ) = opp adj side =. We can then find the hypotenuse: =, then the opposite side = and the adjacent

8 8 opp + adj = hyp () + () = hyp + 9 = hyp hyp = 0 hyp = ± 0 The hypotenuse is always positive so the hypotenuse is 0. sin(θ R ) = opp hyp = = 0 and cos(θ R ) = adj 0 0 hyp = = Since θ is in quadrant III, both the sine and cosine are negative, so sin(θ) = 0 0 cos(θ) = 0 csc(θ) = sec(θ) = cot(θ) = 0 sin(θ) cos(θ) = = tan(θ) = = = = 0 0 = 0 0 = 0 0