MAT 172 SP24 Unit 3 Test Key A

Name: Key A MAT 172 Unit 3 Test: Trig Equations and Identities. Directions: Answer each question and show all of your work. Box in the final solution(s) for each question. GOOD LUCK! 1) Using either angle sum/difference or half-angle, find the measure of each, and be sure any denominators are rationalized: (40 points for each of these) a) sin(285 deg) sin(225 deg + 60 deg) = sin(225 deg) * cos(60 deg) + cos(225 deg) * sin(60 deg) (- sqrt(2))/2 * (1/2) + ((- sqrt(2))/2) * (sqrt(3))/2 = (sqrt(2) * sqrt(6))/4 b) cos((13pi)/12) = cos((10pi)/12 + (3pi)/12) cos((5pi)/6) * cos(pi/4) - sin((5pi)/6) * sin(pi/4) (- sqrt(3))/2 * ((sqrt(2))/2) - 1/2 * ((sqrt(2))/2) = (- sqrt(6) - sqrt(2))/4 sin(570/2) = plus/minus sqrt((1 - cos(520))/2) =- sqrt 1-( - sqrt 3 3 2 )= - sqrt((2 + sqrt(3))/2) = - sqrt(2pi * sqrt(5)) * 2 (sqrt(2) * (sqrt(3) + 1))/4 2) Solve the sine equation, and x: a) Between 0 <= x <= 2pi 4 = 2csc(2x + (7pi)/4) 2=csc( 2x + u = 2x + (7pi)/4 Csc(4)=2 sin mu = 1/Z u = pi/6 u = (5pi)/6 6 = 2 + 2csc(2x + (7pi)/4) ) 12(2x + (2pi)/4) = pi/6 * k 24x + 2 / pi = 2pi 24x =-19-x = (- 19pi)/24 x = (5pi)/24 x = (29pi)/24 X = Period: Yor (12pi)/12 = (24pi)/24 12(2x + (7pi)/4) = (5pi)/6 / 2 24x + 21pi = 10pi 24x = - 117 x = (- 11pi)/24 x = (13pi)/24 X = (37pi)/24 b) General Solutions: (20 points) X = (5pi)/24 plus/minus pi * K X = (L3pi)/24 plus/minus pi * K 3) Solve the equation by using a Pythagorean Identity. Only give all solutions between overline 0 <= overline theta <= overline 2pi (10 overline 0 points) - 3 = cot^2 (theta) - 3csc(theta) - 3 = es * c ^ 2 * theta - 1 - 3csc(theta) (sin^2 theta)/(sin^2 theta) + (cos^2 theta)/(sin^2 theta) = 1 sin^ 2 1 + cot^2 G = csc^2 sigma u = csc(theta) u ^ 2 - 3u + 2 = 0 (u - 2)(u - 1) = 0 u = 2 u = f sin theta = 1/2 sin theta = 1 pi/6 (5pi)/6 갤 B 2 Period: (2pi)/4 = pi/3 = (24pi)/48 4) Solve the equation, and give the values of x: 7^ 4 + 1/2 * cos(4x + (3pi)/4) = 7 ^ (4 + (sqrt(3))/4) lfloor 0 , 96 pi 48 a) Between 0 <= x <= 2pi 90 points cos(4x + (3pi)/4) = (sqrt(3))/2 u = 4x + (3pi)/4 X = (17pi)/48 x = (13pi)/48 cos alpha = (sqrt(3))/2 x = (4i*pi)/48 X = (37pi)/48 u = pi/6 u = 11pi / 6 X = (65pi)/48 x = (6i*pi)/48 12(4x + (3pi)/4) = 12 * pi/6 12(4x + (3pi)/4) = (11 * (1pi)/6) 48x + 9pi = 2pi 48x + 9pi = 22pi x = (89pi)/48 x = (85y)/45 48x = - 7pi x = (13pi)/48 x = (- 7pi)/48 Solutions: (10 points) x = (D*pi)/48 plus/minus pi/z * K x = (13pi)/48 plus/minus pi/2 * K 5) Write the following sums or differences as products with only positive arguments (20 points each) a) cos(83 deg) + cos(69 deg) 2cos((83 + 69)/2) * cos((83 - 64)/2) 2cos(76 deg) * cos(7 deg) b) sin(17/2 * x) - sin(11/2 * x) 2sin((12/2 - 11/2 * x)/2 * x) * cos((17 + 11)/2 * x) 2sin(3/2 * x) * cos(7x) 6) Write the following products as sums or differences with only positive arguments (20 points each) a) - 4cos(9x) * cos(2x) b) -2* sin (36 deg) * cos(61 deg) - 4 * cos(9x + 2x) + cos(9x + 2t) * 2 - 2[cos(7x) + cos(11x)] 2.f sin (3646) + sin(36-6) - [sin(97 deg) + sin(- 25 deg)] sin(97 deg) * sin(25 deg) ] 7) Write the exact value of each, using either angle sum, angle difference, or sum or product rules: (30 points each) a) sin((7pi)/24) * cos(pi/8) - cos((7pi)/24) * sin(pi/8) = 5in((7pi)/24 - pi/5) sin(pi / 6) ne 1 2 b) cos(83 deg) * cos(308 deg) + sin(308 deg) * sin(83 deg) cos(308' - 83') = cos(225 deg) = - (sqrt(2))/2 8) Given that a is in the 1st quadrant and that cos(alpha) = 16/(65') and that ẞ is in the 2nd quadrant and that sin(beta) = 15/(17') please find the following (40 points each) and give the fractions. sin alpha = 63/65 SinB=15/17 cos alpha = 16/65 cos beta = - 8/17 tan alpha = 63/16 tan beta = - 15/8 a) Angles a and ẞ to the nearest tenth of a degree 65 216 alpha = 75.8 deg 15 17 B beta = 118.1 deg b) sin(alpha + beta) sindcos + cosasin B 63/65 * (- 8/17) + 16/65 * (15/17) = - 264/1105 c) sin(alpha - beta) Sindcos B-cos & sing 63/65 * (- 8/17) - 16/65 * (15/17) = - 744/1105 d) cos(alpha - beta) Cos & cosB + sind sin B 16/65 * (- 8/17) + 63/65 * (15/17) = 817/1105 e) cos(alpha/2) give in simplest radical form) 16 Cos (2/2) = (2) = + = tcos(2) 2 65 2 sqrt(81/45) = sqrt(81/130) = (sqrt(9))/(sqrt(130)) = 9 sqrt 130 . sqrt 130 sqrt 130 (9sqrt(130))/130 f) cos(2beta) cos^2 beta - sin^2 beta (- 8/17) ^ 2 - (15/17) ^ 2 = - 161/289 g) sin(2alpha) 2sin alpha * cos alpha 2(63/65)(16/65) = 2016/4225 tan(alpha - beta) h) -744/1105 817/1105 -744/817 i) tan(2beta) (2tan beta)/(1 - tan^2 beta) = 2(-|5| 8 ) 1-(^ 15 6)^ 2 = 240 16| j) csc(alpha + beta) -1105 264