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# Assignment 4 Solutions.pdf

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Department
Calculus
Course Code
Calculus 1000A/B
Professor
Chris Brandl

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MATH 127 Fall 2012 Assignment 4 Solutions 1. Find the exact values. ▯ ▯ ▯▯ ▯1 3 (a) tan cos 5 ▯ ▯ ▯1 3 3 3 Let ▯ = cos . From this we get that cos(▯) since ▯1 ▯5 ▯ 1. This 5 5 corresponds to a right angle trian▯le wi▯ ▯▯djacent side 3, hypotenuse 5, and thus, ▯1 3 4 an opposite side 4. Therefore, tan(▯)tan cos = . ▯ ▯ ▯▯ 5 3 1 (b) sin sin1 p ▯ ▯ e▯▯ 1 1 sin sin1 p = p since these functions are inverses of each other. e e ▯1 (c) cot(cos (x)) ▯1 Let ▯ = cos(x). So then cos(▯) = x which corresponds to a right angle triangle p 2 with adjacent side x, hypotenuse 1, and therefore oppos1 ▯ x . Hence, ▯1 x cot(▯) = cot(cos(x)) =p 2. ▯ ▯ ▯▯ 1 ▯ x 17▯ (d) sin1 sin 6 17▯ Although these functions are inverses of each other this will not cancel out to 17▯ 6 since is not in the range of arcsin(x). We only hav(sin(x)) = x for 6 ▯▯ ▯ x ▯ ▯ but since 2 2 ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ sin 17▯ = sin 5▯ + 2▯ = sin 5▯ = sin ▯ 6 6 6 6 ▯ ▯ ▯▯ ▯1 17▯ ▯1▯ ▯ ▯▯▯ ▯ we get that sin sin = sin sin = . 6 6 6 2. Solve the following equations for x. 1 (a) sin(2x) = cos(x) sin(2x) = cos(x) 2sin(x)cos(x) = cos(x) 2sin(x)cos(x) ▯ cos(x) = 0 cos(x)(2sin(x) ▯ 1) = 0 From this we see that cos(x) = 0 or 2sin(x) ▯ 1 = 0. The ▯rst equation gives ▯ 1 that x = + n▯, n 2 Z. The second equation can be re-written as sin(x) = 2 2 ▯ 5▯ which gives x = 6 + 2n▯ or x = 6 + 2n▯, n 2 Z. ▯ ▯ (b) cos(2x) ▯ cos(x) = ▯ sin (x) + 1 4 ▯ ▯ 1 cos(2x) ▯ cos(x) = ▯ sin (x) + 4 1 cos (x) ▯ sin (x) ▯ cos(x) = ▯sin (x) ▯ 4 2 1 cos (x) ▯ cos(x) + = 0 4 ▯his is a qu▯dratic in terms of cos(x). The quadratic formula gives 1 2 1 cos(x) ▯ = 0. From this we see that cos(x) = and so, 2 2 x = ▯ + 2n▯;▯ ▯ + 2n▯. 3 3 (c) 2sin (x) + sin(x) = 3sin (x) 3 2 2sin (x) + sin(x) = 3sin (x) 2sin (x) ▯ 3sin (x) + sin(x) = 0 2 sin(x)(2sin (x) ▯ 3sin(x) + 1) = 0 sin(x)(2sin(x) ▯ 1)(sin(x) ▯ 1) = 0 1 From this we get sin(x) = 0 or sin(x) = or sin(x) = 1. From these three 2 2 ▯ 5▯ ▯ equations we get x = n▯, x+2n▯; +2n▯, and x = +2n▯ where n 2 Z. 6 6 2 p p (d) 2sin(x) ▯ 2 3cos(x) 3tan(x) + 3 = 0 p p 2sin(x) ▯ 2 3cos(x) ▯ 3tan(x) + 3 = 0 p p sin(x) cos(x) 2(sin(x) ▯3cos(x)) ▯ 3cos(x)+ 3cos(x)= 0 ▯ ▯ p p sin(x) p cos(x) 2(sin(x) ▯ 3cos(x)) ▯3 cos(x)▯ 3 cos(x) = 0 p ! p p sin(x) ▯ 3cos(x) 2(sin(x) ▯ 3cos(x)) ▯ 3 = 0 cos(x) p ! p 3 (sin(x) ▯3cos(x)) 2 ▯ = 0 cos(x) p p From this we see that sin(3cos(x) = 0 or 2cos (x). The ▯rst equation p ▯ p gives tan(x) =3 and so, x = + n▯. The second equation gives cos2x) = 3 and so, x = + 2n▯;11▯+ 2n▯. 6 6 3 ▯ ▯ ▯ 3. Sketch the graph of f(x) = 3cos 2x + 4 over the interval ▯▯ ▯ x ▯ ▯. ▯ ▯ ▯ ▯ ▯▯ ▯ ▯ Re-writing this as f(x) = 3cos 2x + = 3cos 2 x + we see that this graph 4 8 will be a vertical expansion by a factor of 3, a horizontal compression by a factor of 2, followed by a horizontal translation by ▯ to the left. 8 3 2.5 __ (0,3/√ 2 ) 2 1.5 1 0.5 (π/8,0) 3π/8)0 (5π/8)0 0 (-7π/8,0) -0.5 -1 -1.5 -2 -2.5 -3 4. Evaluate the following limits if they exist. If the limit does not exist, show why. 2 x ▯ x ▯ 2 (a) lim 2 x!2 x + 2x ▯ 8 x ▯ x ▯ 2 (x ▯ 2)(x + 1) lim 2 = lim x!2 x + 2x ▯ 8 x!2 (x + 4)(x ▯ 2) x + 1 = lim
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