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You need to refresh all your trig identities. There are plenty of web sites which talk about those. I have high lighted the trig identities that I used in each problem. 1. Find the exact value of the sine, cosine, and tangent of the angle. 11pie/12 Use : sin x = sin (-x) Sin (11/12) = sin(-11/12) = sin(/12) Since we know what is sin pi/6 using the following identity makes sense. Use: cos 2x = 1 – 2sin2 x sinx = sqrt {½ (1-cos2x)} Sin (11/12) = sin(-11/12) = sin(/12) = sqrt {½ (1-cos(/6)}= (1/2 (1-3/2)) Sin (11/12 = (2-3)/2 This is the exact value. Evaluating we get Sin (11/12) = 0.258819 2. Find all solutions of the equation in the interval [0, 2pie). Sin(x-5pie/6)-sin(x+ 5pie/6) =1 Use the trig identity : sin C – sin D = 2 cos (C+D)/2 * sin (C-D)/2 Sin(x-5pie/6)-sin(x+ 5pie/6) = 2 cos x * sin –(5pi/6) = 1 2 cos x * sin –(5pi/6) = 1 cos x * sin 5pi/6 = - 1/2, we know sin 5pi/6 = sin pi – 5pi/6 = sin pi/6 = ½ cos x * 1/2 = -1/2, cos x = -1 Between 0 and 2pi we have only one solution which is x = pi. 3. Find the exact value of sin2x using the double angle formula. Please help explain the use of the double angle formula. Sin x=1/7, 0<x<pie/2 Double angle formula : sin 2x = 2 sin x * cos x Sinx = 1/ 7 and cos x = (1-sin2 x) = ( 1- 1/49) = (48) / 7 = 43/7 So, sin 2x = 2 sin x * cos x = 2 * 1/7 * 43/7 = 83/49 Sin 2x = 83/49 (exactly) Evaluating sin 2x = 0.282784 4. Given cos theta = 4/9, where 3pie/2 <or equal to theta <or equal to 2pie. Question incomplete?? 5. Express 2sin3xcos6x as a sum containing only sines or cosines. There are 4 identities involving sin C sin D and cos C cos D. These are very useful. We will use one of them. (note we used one in prob 2) Identity used: sin C + sin D = 2 sin (C+D)/2 * cos (C-D)/2 2sin3xcos6x = 2 sin (C+D)/2 * cos (C-D)/2 ( we will find C and D) (C+D)/2 = 3x C+D = 6x, (C-D)/2 = 6x C-D = 12 x Solving for C and D we get C = 9x, D = -3x So we can write 2sin3xcos6x = sin 9x + sin (-3x ) = sin 9x + (-sin 3x) 2sin3xcos6x = sin 9x + sin (-3x ) = sin 9x + (-sin 3x) 6. Express cos7x-cos5x as a product containing only sines and/or cosines. Similar to 5. Identity used: cos C – cos D = 2 sin (C+D)/2 * sin (D-C)/2 cos7x-cos5x = 2 sin(6x)sin(-x) = -2 sinx sin 6x 7. Evaluate the expressions without the aid of a calculator. a. Arctan(- sqrt3/3) b. Arcsin(-1/2) Assume x = tan –1 (-3/3) = tan –1 -(1/3) tan x = -1/3 tan 2 x = 1/3 use the fact that 1 + tan 2 x = sec 2 x sec 2 x = 1 + 1/3 = 4/3 sec x = 1/cos x = 2 / 3 cos x = 3 / 2 (familiar ?) positive solution : cos x = 3 / 2 x= pi/6 cosine is positive in the first and forth quadrant. But we are interested only a solution in the 2nd and 4th quadrant because tan is negative in these quadrants. (If you see the original expression the given inverse tan function has a negative value and tan function are negative only in the 2nd and 4th quads ) So one solution is x = 2pi -pi/6 = 11 pi/6 negative solution : cos x = -3 / 2 cosine is negative in the 2nd and 3rd quadrant. So solution could be x = pi – pi/6 and/or pi + pi/6, ie. X = 5pi/6, 7pi/6. But we are interested only a solution in the 2nd and 4th quadrant Other solution is 5pi/6 Arctan(- sqrt3/3) = 5pi/6 or 11pi/6 (b) Arcsin(-1/2) = x Sin x = -1/2 x = -pi/6 Sin is negative in 3rd and 4th quads. So x = pi+pi/6 = 7pi/6 and X = 2pi – pi/6 = 11pi/6 Arcsin(-1/2) = 7pi/6 or 11pi/6 8. Use inverse functions to evaluate the expressions, a. cos(arcsin(sqrt5/5)) , say x= arcsin(sqrt5/5)) sin x = sqrt5/5 sin x = 1/ 5 cos(arcsin(sqrt5/5)) = cos x = ? Use a right triangle to find cosx which is what we need. sin x = 1/ 5 = opposite side / hypotenuse. Cos x = adjacent /hypo. = 2/5 b. cos(arctan(sqrt2/x)) As before, say, r = arctan(sqrt2/x)) tan r = 2/x, hypot of the right triangle = (x2 + 2) So cos(arctan(sqrt2/x)) = cos r adjacent/ hypot. = x/(x2 + 2) 9. = Identify he x-values that are solutions of the equations. a. 8cos x-4 = 0 b. 18cot^2 x-18 = 0 8cos x-4 = 0 angles. cos x = ½, x = pi/3, 2pi- pi/3 = 5pi/3, cosine is cyclic in 2pi General solutions are x = pi/3 2npi and x = 5pi/3 2npi where n = integer 18cot^2 x-18 = 0 cot^2 x = 1 tan^2 x = 1 sec2 x = 1 +tan^2 x = 2 cos^2 x = ½ cos x = ½ positive: cos x = ½ similar to part (a) Solution set 1: x = pi/3 2npi and x = 5pi/3 2npi where n = integer negative cos x = -1/2 x = pi pi/3 Solution set 2: 4pi/3 2n pi or 2pi/3 2n pi , n= integer 10. A large pole is 175 feet tall. On a particular day at noon it casts a 198-foot shadow. What is the sun’s angle of elevation? Tan x = 175/198 Elevation = 41.5o