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Calculus 1000A/B
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Chris Brandl
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Lecture

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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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