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Lecture

Calculus 1000A/B Lecture Notes - Trigonometry, Hypotenuse, Scilab


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

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MATH 127 Fall 2012
Assignment 4 Solutions
1. Find the exact values.
(a) tan cos13
5
Let θ= cos13
5. From this we get that cos (θ) = 3
5since 13
51. This
corresponds to a right angle triangle with adjacent side 3, hypotenuse 5, and thus,
an opposite side 4. Therefore, tan (θ) tan cos13
5=4
3.
(b) sin sin11
e
sin sin11
e=1
esince these functions are inverses of each other.
(c) cot (cos1(x))
Let θ= cos1(x). So then cos(θ) = xwhich corresponds to a right angle triangle
with adjacent side x, hypotenuse 1, and therefore opposite side 1x2. Hence,
cot (θ) = cot (cos1(x)) = x
1x2.
(d) sin1sin 17π
6
Although these functions are inverses of each other this will not cancel out to 17π
6
since 17π
6is not in the range of arcsin(x). We only have sin1(sin(x)) = xfor
π
2xπ
2but since
sin 17π
6= sin 5π
6+ 2π= sin 5π
6= sin π
6
we get that sin1sin 17π
6= sin1sin π
6=π
6.
2. Solve the following equations for x.
1

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(a) sin (2x) = cos (x)
sin (2x) = cos (x)
2 sin (x) cos (x) = cos (x)
2 sin (x) cos (x)cos (x)=0
cos (x)(2 sin (x)1) = 0
From this we see that cos (x) = 0 or 2 sin (x)1 = 0. The first equation gives
that x=π
2+,nZ. The second equation can be re-written as sin (x) = 1
2
which gives x=π
6+ 2or x=5π
6+ 2,nZ.
(b) cos (2x)cos (x) = sin2(x) + 1
4
cos (2x)cos (x) = sin2(x) + 1
4
cos2(x)sin2(x)cos (x) = sin2(x)1
4
cos2(x)cos (x) + 1
4= 0
This is a quadratic in terms of cos (x). The quadratic formula gives
cos (x)1
22
= 0. From this we see that cos (x) = 1
2and so,
x=π
3+ 2, π
3+ 2.
(c) 2 sin3(x) + sin (x) = 3 sin2(x)
2 sin3(x) + sin (x) = 3 sin2(x)
2 sin3(x)3 sin2(x) + sin (x)=0
sin (x)(2 sin2(x)3 sin (x) + 1) = 0
sin (x)(2 sin (x)1)(sin (x)1) = 0
From this we get sin (x) = 0 or sin (x) = 1
2or sin (x) = 1. From these three
2
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equations we get x=,x=π
6+ 2, 5π
6+ 2, and x=π
2+ 2where nZ.
(d) 2 sin (x)23 cos (x)3 tan (x) + 3 = 0
2 sin (x)23 cos (x)3 tan (x) + 3 = 0
2(sin (x)3 cos (x)) 3sin (x)
cos (x)+ 3cos (x)
cos (x)= 0
2(sin (x)3 cos (x)) 3sin (x)
cos (x)3cos (x)
cos (x)= 0
2(sin (x)3 cos (x)) 3 sin (x)3 cos (x)
cos (x)!= 0
(sin (x)3 cos (x)) 23
cos (x)!= 0
From this we see that sin (x)3 cos (x) = 0 or 23
cos (x)= 0. The first equation
gives tan (x) = 3 and so, x=π
3+. The second equation gives cos (x) = 3
2
and so, x=π
6+ 2, 11π
6+ 2.
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