Class Notes (1,100,000)

CA (650,000)

Western (60,000)

CALC (700)

CALC 1000A/B (500)

Chris Brandl (2)

Lecture

This

**preview**shows pages 1-2. to view the full**8 pages of the document.**MATH 127 Fall 2012

Assignment 4 Solutions

1. Find the exact values.

(a) tan cos−13

5

Let θ= cos−13

5. From this we get that cos (θ) = 3

5since −1≤3

5≤1. This

corresponds to a right angle triangle with adjacent side 3, hypotenuse 5, and thus,

an opposite side 4. Therefore, tan (θ) tan cos−13

5=4

3.

(b) sin sin−11

√e

sin sin−11

√e=1

√esince these functions are inverses of each other.

(c) cot (cos−1(x))

Let θ= cos−1(x). So then cos(θ) = xwhich corresponds to a right angle triangle

with adjacent side x, hypotenuse 1, and therefore opposite side √1−x2. Hence,

cot (θ) = cot (cos−1(x)) = x

√1−x2.

(d) sin−1sin 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 sin−1(sin(x)) = xfor

−π

2≤x≤π

2but since

sin 17π

6= sin 5π

6+ 2π= sin 5π

6= sin π

6

we get that sin−1sin 17π

6= sin−1sin π

6=π

6.

2. Solve the following equations for x.

1

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

(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+nπ,n∈Z. The second equation can be re-written as sin (x) = 1

2

which gives x=π

6+ 2nπ or x=5π

6+ 2nπ,n∈Z.

(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+ 2nπ, −π

3+ 2nπ.

(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

###### You're Reading a Preview

Unlock to view full version

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

equations we get x=nπ,x=π

6+ 2nπ, 5π

6+ 2nπ, and x=π

2+ 2nπ where n∈Z.

(d) 2 sin (x)−2√3 cos (x)−√3 tan (x) + 3 = 0

2 sin (x)−2√3 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)) 2−√3

cos (x)!= 0

From this we see that sin (x)−√3 cos (x) = 0 or 2−√3

cos (x)= 0. The ﬁrst equation

gives tan (x) = √3 and so, x=π

3+nπ. The second equation gives cos (x) = √3

2

and so, x=π

6+ 2nπ, 11π

6+ 2nπ.

3

###### You're Reading a Preview

Unlock to view full version