Assignment query 5

course mth164

?G?y????????????assignment #005005.

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

Precalculus II

06-10-2007

......!!!!!!!!...................................

12:17:55

Query problem 6.2.8 exact value of tan(195 deg)

......!!!!!!!!...................................

RESPONSE -->

.2679491924

confidence assessment: 1

.................................................

......!!!!!!!!...................................

12:18:12

what is the exact value of tan(195 deg)?

......!!!!!!!!...................................

RESPONSE -->

.2679491924

confidence assessment: 1

.................................................

......!!!!!!!!...................................

12:19:07

to get this result you used the exact values of two angles to get tan(15 deg). Which two angles were these, what were the exact values, and what formula did you use to get your result?

......!!!!!!!!...................................

RESPONSE -->

tangent = sin/cos

tangent(15) = sin(15)/cos(15)

confidence assessment: 14

**None of your results gives an exact value. Everything you did here would result in a decimal approximation.

To get the exact value:

tan(alpha+beta)= [tan(alpha)+tan(beta)]/[1-tan(alpha)tan(beta].

So tan(195 deg) = tan(150 deg + 45 deg) =

[ tan(150 deg) + tan(45 deg) ] / [ 1 - tan(150 deg) * tan(45 deg) ] =

[ -sqrt(3) / 3 + 1 ] / [ 1 - (-sqrt(3)) * 1 ] =

[1 - sqrt(3)/3] / (1 + sqrt(3)/3 )=

(1 - sqrt(3)/3)(1 - sqrt(3)/3) / [ (1+sqrt(3)/3)(1-sqrt(3)/3) ] =

(1 - 2 sqrt(3)/3 + 1/3) / (1 - 1/3) =

(4/3 - 2 sqrt(3)/3) / (2/3) =

2 - sqrt(3). **

.................................................

......!!!!!!!!...................................

12:20:50

For which angles between 0 and 90 degrees do you know the exact value of the trigonometric functions? How many ways are there to combine angles you have listed to obtain 15 degrees?

......!!!!!!!!...................................

RESPONSE -->

0,30,45,60,90

45-30

60-45

confidence assessment: 1

.................................................

......!!!!!!!!...................................

12:21:30

Query problem 6.2.20 cos(5 `pi / 12) cos(7`pi/12) - sin(5`pi/12)sin(7`pi/12)

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:24:42

what is the exact value of the expression?

......!!!!!!!!...................................

RESPONSE -->

Using the identity for the Cosine of a sum we can rewrite the problem as follows:

cos(5pi/12)cos(7pi/12)-sin(5pi/12)sin(7pi/12)=

cos(5pi/12+7pi/12)=

cos(12pi/12)=

cos(pi) = -1

confidence assessment: 3

Good.

To specifically answer the question:

** We use the identity cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta), with alpha = 5 pi/12 and beta = 7 pi / 12:

cos(5 `pi / 12) cos(7`pi/12) - sin(5`pi/12)sin(7`pi/12) = cos(12 `pi/ 12) - sin(12`pi/ 12), so our equation becomes

cos(`pi) - sin(`pi) = cos(`pi) - sin(`pi)

which is an identity. If we wish we can evaluate these expressions to get

-1 - 0 = -1 = 0 or

-1 = -1. **

.................................................

......!!!!!!!!...................................

12:24:59

How did you obtain your result?

......!!!!!!!!...................................

RESPONSE -->

I used the identiy for the sum of cosines

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:25:08

Which sum or difference formula does this expression fit?

......!!!!!!!!...................................

RESPONSE -->

the sum of cosines

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:25:35

If this expression is the right-hand side of a sum or difference formula, what then is the left-hand side and what is its exact value?

......!!!!!!!!...................................

RESPONSE -->

the left side is cos(5pi/12+7pi/12)=-1

confidence assessment: 3

Good.

.................................................

......!!!!!!!!...................................

12:25:57

Query problem 6.2.38 tan(2 `pi - `theta) = - tan(`theta)

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:26:46

how do you establish the given identity?

......!!!!!!!!...................................

RESPONSE -->

It looks like a sum formula for tangent

confidence assessment: 3

** Using the identity

tan(alpha-beta) = (tan(alpha) - tan(beta) ) / (1 + tan(alpha) * tan(beta) ).

If alpha = 2 `pi and beta = theta then we have

tan(2 pi - theta)= (tan 2pi-tan theta)/(1+tan(2pi) * tan (theta) ).

We know that the value of tan of 2pi =0 so we can say

tan(2 pi - theta)= (0-tan(theta))/(1+(0)(tan (theta) ) = (- tan (theta))/1

or

tan(2 pi - theta)= - tan (theta) **

** Alternative (and much simpler) solution:

We can use the periodicity and even/odd properties of sine and cosine.

sin(-`theta) = - sin(`theta), cos(-`theta) = - cos(`theta) so

tan (-`theta) = - tan(`theta).

All functions have the same values at 2 `pi + `theta as at `theta, so the result follows. **

.................................................

......!!!!!!!!...................................

12:36:24

Query problem 6.3.8 find exact values of sin & cos of 2`theta, and of `theta/2 if csc(`theta) = -`sqrt(5)

......!!!!!!!!...................................

RESPONSE -->

since csc=1/sin then sin(theta)=-1/sqrt(5)

so using cos(2theta)=1-2(sin(theta))^2

cos(2theta)=1-2(-1/sqrt(5))^2=3/5

sin(2theta)=2sin(theta)cos(theta)=(-1/sqrt(5))(2/sqrt(5))

= 2/5 (positive or negative)

theta/2 would be about -.2318 radians

confidence assessment: 2

`aGood. More detail follows:

** Since csc(theta) = 1 / sin(theta) we know that sin(theta) = 1 / csc(theta) so we have

sin(theta) = 1 / csc(theta) = 1 / (-sqrt(5) ) = - sqrt(5) / 5.

We also know from sin^2(theta) + cos^2(theta) = 1 that cos(theta) = sqrt(1 - sin^2(theta) ) so we have

cos(theta) = +-sqrt( 1 - ( -sqrt(5) / 5)^2 ) = +-sqrt( 1 - 5/25) = +-sqrt(20/25) = +-sqrt( 5 * 4 / 25) = +-sqrt(4/25) * sqrt(5) = +-2 * sqrt(5) / 5.

The +- depends on whether theta is in the third or fourth quadrant, in both of which the sine is negative. For a third-quadrant angle the cosine is negative and in the fourth it is positive.

Now:

sin(2`theta) = 2 sin(`theta) cos(`theta) = 2 (-sqrt(5) / 5) * +-2 * sqrt(5) / 5 = +-4 * 5 / 25 = +-4/5.

cos(2`theta) = cos^2(`theta) - sin^2(`theta) = (+-2 sqrt(5) / 5)^2 - (-sqrt(5)/5)^2 = 20/25 - 5/25 = 15/25. **

sin(`theta/2) = +- `sqrt( (1 - cos(`theta) ) / 2) = +- sqrt( 1 - (+- 2 sqrt(5) / 5) ) = +- sqrt( 1 -+ sqrt(5)/5).

cos(`theta) = +-`sqrt( (1 + cos(`theta)) / 2) = +- sqrt( 1 + (+- 2 sqrt(5) / 5) ) = +- sqrt( 1 +- sqrt(5)/5). **

.................................................

......!!!!!!!!...................................

12:36:38

what are the exact values of the given expressions?

......!!!!!!!!...................................

RESPONSE -->

answered in previous answer

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:36:57

Which formulas did you used to obtain these results?

......!!!!!!!!...................................

RESPONSE -->

Used double angle formula and formula for csc

confidence assessment: 2

.................................................

......!!!!!!!!...................................

12:38:12

Explain how the identities sin(2`theta) = 2 sin(`theta) cos(`theta), cos(2`theta) = cos^2(`theta) - sin^2(`theta), sin(`alpha/2) = +- `sqrt( (1 - cos(`alpha) ) / 2), cos(`alpha) = +-`sqrt( (1 + cos(`alpha) ) / 2) are used to get your results.

......!!!!!!!!...................................

RESPONSE -->

I could easily find sin(theta) using the fact that csc=1/sin. So I then found cos using pythagoreans identity then plugged the results into a double angle formula

confidence assessment: 2

.................................................

......!!!!!!!!...................................

12:38:21

Query problem 6.3.20 exact value of csc(7`pi/8)

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:43:20

what is the exact value the given expression?

......!!!!!!!!...................................

RESPONSE -->

csc(7pi/8)=csc(.5(7pi/4))=1/sin(.5(7pi/4)

using half angle forumul

sin(.5(7pi/4))=+/- sqrt((1-cos(7pi/4))/2)=

+/- sqrt((1-sqrt(2)/2)/2)

confidence assessment: 2

Again good. Compare with:

`a** sin(7`pi/8) = sin( 1/2 * 7`pi/4). You know the exact value of sin(7`pi/4).

Use the half-angle formula. You get sin(7 pi / 8) = `sqrt(2`sqrt(2)+4) ).

Thus csc( 7 pi / 8) = 1 / sin(theta) = 1 / sqrt( 2 sqrt(2) + 4)). **

.................................................

......!!!!!!!!...................................

12:44:20

Explain how you obtained your result from the half-angle formulas.

......!!!!!!!!...................................

RESPONSE -->

I rewrote 7pi/8 as (1/2)(7pi/4) and since I knew the exact value of cos(7i/4) then substituting into the half angle formula I arrived at the exact answer

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:44:37

query problems 6.3.42 tan(`theta/2) = csc(`theta) - cot(`theta)

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:46:00

explain how you established the given identity

......!!!!!!!!...................................

RESPONSE -->

rewrite the left side using the half angle identity for tan. Rewrite the right side using the fundamental identites for csc and cot. Multiplying both sides by the common denominator of sin(theta) yields 1-cos(theta) on both sides of the equal sign

confidence assessment: 3

.................................................

......!!!!!!!!...................................

12:46:39

Comm on any surprises or insights you experienced as a result of this assignment.

......!!!!!!!!...................................

RESPONSE -->

I accidentally skipped one problem and misread another. But now I understand where the problems were leading to.

confidence assessment: 2

.................................................

"

&#

This looks good. See my notes. Let me know if you have any questions. &#