Assignment 7 Query

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course Mth 164

3/22 4:30

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

Precalculus II

Asst # 7

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Self-critique (if necessary):

OK

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Self-critique Rating: OK

**** Query problem 6.6.12 cot(2`theta/3) = -`sqrt(3)

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Your solution:

If we have the equation cot(2`theta/3) = -`sqrt(3), then tan(2`theta/3) = -`sqrt(3)/3. In the interval [0, pi), the tangent function has a value of -`sqrt(3)/3 when the argument is 5pi/6, therefore, the general formula that gives all the solutions is theta = 5pi/4 + 6pi/4 k where k = any integer and this is figured as follows:

2`theta/3 = 5pi/6 + pi k

2`theta = 15pi/6 +3pi k

`theta = 15pi/12 + 3pi/2 k

`theta = 5pi/4 + 6pi/4 k

So, for the interval [0, 2pi) we can find the solution as follows:

`theta = 5pi/4 + 6pi/4 * 0 = 5pi/4

The solution set is {5pi/4}.

confidence rating #$&*: 3

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Given Solution:

The cotangent function takes value -sqrt(3) when its argument is 5 pi / 6 or 11 pi / 6, as can easily be seen using a labeled unit circle and the definition of the cotangent.

If the argument of the function is coterminal with 5 pi / 6 or 11 pi / 6 we obtain the same value for the cotangent.

So the possible arguments are 5 pi / 6 + 2 pi n or 11 pi / 6 + 2 pi n, where n can be any integer.

The student solution given below is correct for arguments 5 pi / 6 and 11 pi / 6, but does not address possible coterminal arguments.

2`theta/3 = 5pi/6 or 11pi/6

`theta = 5pi/6 * 3/2 or 11pi/6 * 3/2

`theta = 5pi/4 or 11pi/4

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23:43:44

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Self-critique (if necessary):

OK

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Self-critique Rating: OK

**** List all the possible values of 2 `theta / 3 such that the equation is satisfied (the list is infinite; use the ellipsis ... to indicate the continuation of a pattern)

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Your solution:

theta = 5pi/4 + 6pi/4 k so,

all the possible values of 2`theta/3 is:

(2 * (5pi/4 + 6pi/4 k))/3 where k = 1,2,3,4…for example:

cot(2 * (5pi/4 + 6pi/4 * 0))/3 = cot(2 * (5pi/4))/3 = -`sqrt(3)

cot(2 * (5pi/4 + 6pi/4 * 1))/3 = cot(2 * (11pi/4))/3 = -`sqrt(3)

cot(2 * (5pi/4 + 6pi/4 * 2))/3 = cot(2 * (17pi/4))/3 = -`sqrt(3)

cot(2 * (5pi/4 + 6pi/4 * 3))/3 = cot(2 * (23pi/4))/3 = -`sqrt(3)…

confidence rating #$&*: 3

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Given Solution:

2`theta/3 = 2/3(5pi/4 + kpi)

when k=0 2/3`theta = 5pi/6

when k=1 2/3`theta = 11pi/6

when k=2 2/3`theta = 17pi/6

when k=3 2/3`theta = 235pi/6

when k=4 2/3`theta = 29pi/6

when k=5 2/3`theta = 35pi/6

when k=6 2/3`theta = 41pi/6

etc ........

when k=n 2/3`theta = 5pi/6 + npi

** 2 `theta / 3 can be 5 `pi / 6 or 5 `pi / 6 + 2 `pi or 5 `pi / 6 + 4 `pi or in general 5 `pi / 6 + 2 n `pi, or it can be

11 `pi / 6 or 11 `pi / 6 + 2 `pi or 11 `pi / 6 + 4 `pi or in general 11 `pi / 6 + 2 n `pi. **

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23:57:51

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Self-critique (if necessary):

After studying the solution, I don’t believe that understood the question.

???What does “all the values of 2`theta/3” mean?

I understand that 2`theta/3 = 5pi/6 + pi k where k is any integer.

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Self-critique Rating: 1

@& 2 theta / 3 can be any angle which differs from 5 pi / 6 by an integer multiple of pi.*@

**** How many of these values result in `theta values between 0 and 2 `pi?

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23:59:12

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Your solution:

Only one of these values result in `theta values between 0 and 2 `pi.

`theta = 5pi/4 + 6pi/4 * -1 = -pi/4 is < 0

`theta = 5pi/4 + 6pi/4 * 0 = 5pi/4 is between 0 and 2pi

`theta = 5pi/4 + 6pi/4 * 1 = 11pi/4 is > 2pi

confidence rating #$&*: 3

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Given Solution:

** If 2 `theta / 3 = 5 `pi / 6 + 2n `pi then multiplying both sides by 3/2 we get

`theta = 5 `pi / 2 + 3 n `pi. Since 3 n `pi > 2 `pi only 5 `pi / 2 is between 0 and 2 `pi.

If 2 `theta / 3 = 11 `pi / 6 + 2n `pi then multiplying both sides by 3/2 we get

`theta = 11 `pi / 2 + 3 n `pi. Since 11 `pi / 2 > 2 `pi this can't lie between 0 and 2 `pi unless n is negative. **

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Self-critique (if necessary):

I don’t think this solution is correct.

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Self-critique Rating: 3

@& You're right. The solution should read

If 2 theta / 3 = 5 pi / 6 + n * pi then multiplying by 3/2 we get

theta = 5 pi / 4 + 3/2 n pi.

For negative values of n we would get negative values of theta, which are not between 0 and 2 pi.

For n = 0 we get theta = 5 pi / 4.a

For n = 1 we get theta = 3 pi / 2 + 5 pi / 4, which is greater than 2 pi.*@

**** Query problem 6.6.44 solve sin^2(`theta) = 2 cos(`theta) + 2

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14:41:12

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Your solution:

We have the equation sin^2(`theta) = 2 cos(`theta) + 2 so,

sin^2(`theta) = 2 cos(`theta) + 2

1 - cos^2(`theta) = 2 cos(`theta) + 2

cos^2(`theta) + 2 cos(`theta) + 2 = 1

cos^2(`theta) + 2 cos(`theta) + 1 = 0

(cos(theta) +1)(cos(theta) +1) = 0 therefore,

cos(theta) = -1 and cos^-1(-1) = theta = pi

In the interval [0, 2pi), the solution set is {pi}.

All of the solution can be given by using the general form as follows:

{theta| theta = pi + 2pi k, where k is any integer}.

confidence rating #$&*: 3

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Given Solution:

** Since sin^2(`theta) = 1 - cos^2(`theta) we have

1 - cos^2(`theta) = 2 cos(`theta) + 2. This equation is a quadratic equation in cos(`theta). To see this rearrange the equation to get

cos^2(`theta) + 2 cos(`theta) + 1 = 0. Now let u = cos(`theta). You get

u^2 + 2 u + 1 = 0. This is a quadratic equation with solution u = -1.

Thus our solution is u = -1, meaning cos(`theta) = -1, so `theta = -pi + 2 `pi k, for any integer k. **

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14:43:04

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Self-critique (if necessary):

OK

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Self-critique Rating: OK

**** Query problem 6.6.66 19x + 8 cos(x) = 0

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Your solution:

I believe this problem is state in the book as:

19x + 8 cos(x) = 2.

By using a graphing utility to solve this equation, we obtain the solution set {approx. -.30}.

confidence rating #$&*: 3

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Given Solution:

19x+8cos(x)=2 by putting 19x+8 cos(x) in as y sub1 and 2 as y sub 2 we can

trace and find that the closest value for x is approximately (.30).

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Self-critique (if necessary):

This given solution is incorrect.

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Self-critique Rating: 3

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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

*** File was originally submitted on 3/21 or 3/22, was critiqued by failed to post. ***