assignment 7

course mth 164

v?????o????J??assignment #007

007.

Precalculus II

11-11-2008

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22:31:55

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

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RESPONSE -->

cot=tan^-1(2x/3)=-sqrt3

x=

confidence assessment: 0

If you were to sketch the unit circle with all angle multiples of pi/6 and pi/4, and evaluate the trigonometric functions for all these angles, you would find that cot(z) = -sqrt(3) when z = 5 pi/6 or when z = 11 pi/6. (To verify this: At these points the unit circle points have coordinates (-sqrt(3)/2, 1/2) and (sqrt(3)/2, -1/2) so that the cotangent, which is equal to x / y, has values (-sqrt(3)/2) / (1/2) = -sqrt(3) and (sqrt(3)/2)/(-1/2) = -sqrt(3).)

The cotangent has the same values for all z values coterminal with these points. So solutions include z = 5pi/6 + 2 n pi, and z = 11 pi / 6 + 2 n pi .

So cot(2 theta / 3) = -sqrt(3) when 2 theta / 3 = 5 pi /6, and also when 2 theta / 3 = 11 pi / 6.

2 theta / 3 = 5 pi /6 has solution theta = 5 pi / 4.

2 theta / 3 = 11 pi / 6 has solution theta = 11 pi / 4.

Additional solutions are obtained by solving the equations

2 theta / 3 = 5 pi /6 + 2 n pi and

2 theta / 3 = 11 pi / 6 + 2 n pi.

However all these solutions are greater than theta = 2 pi, as is the solution theta = 11 pi / 4. Unless values of theta outside the domain 0 <= theta < 2 pi are desired, the only solution of interest is the solution

theta = 5 pi / 4.

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22:32:28

what are your solutions?

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RESPONSE -->

solutions are in radians and they are corresponding with the unit circle

confidence assessment: 1

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22:33:09

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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RESPONSE -->

this pattern is constantly increasing

confidence assessment: 2

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22:33:45

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

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RESPONSE -->

none

confidence assessment: 1

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22:45:18

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

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RESPONSE -->

pi=x

confidence assessment: 1

** 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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22:47:35

what are your solutions to the equation?

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RESPONSE -->

pi to -pi is the range

confidence assessment: 1

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22:47:55

How did you simplify your equation before obtaining your solutions?

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RESPONSE -->

by using the properties in the book

confidence assessment: 1

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23:02:03

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

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RESPONSE -->

x=0 or pi/2 because of the coordinate value 0,1

confidence assessment: 2

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23:02:18

what is your approximate solution?

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RESPONSE -->

0 or pi/2

confidence assessment: 2

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23:03:19

Explain how used your graphing utility (calculator, computer or whenever) to solve this equation.

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RESPONSE -->

I used the calculator to solve as well as the unit circle because i had to check myself with the calculator

confidence assessment: 2

One way to solve the equation approximately is to graph y = 19 x + 8 cos(x).

The graph crosses the x axis somewhere between x = -.3 and x = -.4.

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23:04:25

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

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RESPONSE -->

I noticed that I need to study my section notes and aslo need to familiarize myself more with the calculators equation solver function(read book)

confidence assessment: 3

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?N?????????d??assignment #007

007.

Precalculus II

11-11-2008

I've inserted notes on every problem. Please be sure you know how to solve problems of this type. Study the text, take another good look at the techniques covered in the qa's, and check out the Class Notes on your disks.

I'll be glad to answer additional questions--just submit them using the Submit Work Form or the Questions form.