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course mth 164
10/23 4pm
Precalculus IIAsst # 7
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Self-critique (if necessary):
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Self-critique Rating:
**** Query problem 6.6.12 cot(2`theta/3) = -`sqrt(3)
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Your solution:
Arccot(cot(2theta/3)) = arccot(-sqrt3)
2theta/3 = 5pi/6 ?
How do I get theta alone in this situation?
confidence rating #$&*: 1
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Given Solution:
The cotangent function takes value -sqrt(3) when its argument is 5 pi / 6 or 11 pi / 6, as can easily be seen using a labeled unit circle and the definition of the cotangent.
If the argument of the function is coterminal with 5 pi / 6 or 11 pi / 6 we obtain the same value for the cotangent.
So the possible arguments are 5 pi / 6 + 2 pi n or 11 pi / 6 + 2 pi n, where n can be any integer.
The student solution given below is correct for arguments 5 pi / 6 and 11 pi / 6, but does not address possible coterminal arguments.
2`theta/3 = 5pi/6 or 11pi/6
`theta = 5pi/6 * 3/2 or 11pi/6 * 3/2
`theta = 5pi/4 or 11pi/4
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23:43:44
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Self-critique (if necessary): So I multiply both sides by 3/2? Confused.
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Self-critique Rating:2
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If you multiply both sides of the equation
2 theta / 3 = 5 pi / 6
by 3, you get
2 theta = 15 pi / 6.
If you then divide by 2 you get
theta = 15 pi / 12,
which reduces to 5 pi / 4.
Alternatively you could just multiply both sides of the original equation by 3/2, which is the multiplicative inverse of 2/3.
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**** List all the possible values of 2 `theta / 3 such that the equation is satisfied (the list is infinite; use the ellipsis ... to indicate the continuation of a pattern)
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Your solution:
2theta/3 = 2/3 (5pi/4 + kpi)
confidence rating #$&*:
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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):I don’t understand how to distribute the k value throughout the equation?
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Self-critique Rating:3
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No wonder you have a question on this one. The given solution does not correspond to the given problem.
Here's a solution to the given problem.
The cotangent function is periodic with period pi. Thus if any integer multiple of pi is added to 2`theta/3, we still have a solution.
So our solutions have the form 2/3 theta + n * pi = 5 pi / 4, where n can be any integer.
We can subtract the n * pi from both sides to get
2/3 theta = 5 pi / 4 - n * pi.
Noting that n can be a positive or negative integer, we get all the same solutions if we change the sign of n, so that our solutions could be written
2/3 theta = 5 pi / 4 + n * pi.
For example, when n = 2 we would get the equation
2/3 theta = 5 pi / 4 + 2 * pi
Writing the right-hand side with common denominator 4 we have
2/3 theta = 5 pi / 4 + 8 pi / 4 = 13 pi / 4
so that
theta = 3/2 * 13 pi / 4 = 39 pi / 8.
We could individually substitute values 0, 1, 2, 3, ... for n, and would obtain the following solutions:
For n = 0, theta = 15 pi / 8.
For n = 1, theta = 27 pi / 8.
For n = 2, theta = 39 pi / 8.
For n = 3, theta = 51 pi / 8.
For n = 4, theta = 63 pi / 8.
For n = 5, theta = 75 pi / 8.
For n = 6, theta = 87 pi / 8.
Alternatively we could solve the equation
2/3 theta = 5 pi / 4 + n * pi
for theta, obtaining
theta = 3/2 ( 5 pi / 4 + n pi) = 3/2 ( 5 pi / 4 + 4 n pi / 4) = 3/2 (5 + 4 n) * pi / 4 = (15 + 12 n pi) / 8.
and substitute n values 0, 1, 2, 3, ... .
We would obtain the same values of theta as before.
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**** 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:
confidence rating #$&*:
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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):
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Self-critique Rating:
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Corrected solution:
From preceding calculations, we see that n = 0 yields theta = 15 pi / 8, which is less than 2 pi. It is also clear from preceding results that no positive value of n gives us a solution between 0 and 2 pi.
We could begin finding solutions for negative values of n, and if so we would quickly see that n = -1 gives us a solution between 0 and 2 pi, but no other negative value of n does so.
A more powerful method, for which we would be grateful if there were a large number of such solutions, is as follows:
We know from before that our solutions are of the form
theta = (15 + 12 n) pi / 8, for integer values of n.
0 <= theta < 2 pi means
0 <= (15 + 12 n ) pi / 8 < 2 pi.
Multiplying all expressions by 8 we get
0 <= (15 + 12 n) pi < 16 pi.
Dividing both sides by pi we have
0 <= 15 + 12 n < 16 so that
-15 <= 12 n < 1 and
-5/4 < n < 1/12.
So n is an integer between -5/4 and 1/12.
The only integers that satisfy this are -1 and 0.
So n = -1 and n = 0 give us values of theta between 0 and 2 pi.
These values are found by substituting 0 and -1 into our solution for theta, obtaining
theta = 15 pi / 8 (for n = 0) and
theta = 3 pi / 8 ( for n = -1).
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**** 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:
confidence rating #$&*:
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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):
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Self-critique Rating:
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&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).
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**** Query problem 6.6.66 19x + 8 cos(x) = 0
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Your solution:
confidence rating #$&*:
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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):
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Self-critique Rating:
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Expanded solution:
The equation
19 x + 8 cos(x) = 2
is easily rearranged to the form
8 cos(x) = 2 - 19 x.
Each side of the equation is an easily-graphed simple function. If the graphs of the two functions intersect, then at the corresponding value of x, the equation is solved.
A graph of 2 - 19 x has slope -19 and y-intercept 2.
A graph of 8 cos(x) has amplitude 8 and y intercept (0, 8). Its first intercept with the positive x axis is at (pi/2, 0).
It is clear with any reasonable hand-sketched graph that these two graphs do not intersect anywhere to the right of the y axis. The y intercept of the linear graph is lower than that of the cosine graph, and the linear graph is far too steep to intersect the cosine function as it moves to the right. By the time x = pi / 2, the value of y = 2 - 19 x is already less than -20, and the lowest point on the cosine graph is at y = -8.
It is equally clear that the graph of the linear function will intersect the cosine graph to the left of the y axis, and that because of its steepness it will do so rather quickly. By the time the cosine graph reaches the point (-pi/2, 0), the graph of y = 2 - 19 x is greater than 30, far exceeding the peak value 8 of the cosine function. The linear function reaches value 8 at approximately x = -.32, and must therefore have intersected the graph of the cosine function before reaching value 8. A reasonable estimate of the x value of the intersection point would therefore be about -.3.
Using a graphing calculator it is easy to verify that 19 x + 8 cos(x) = 2 occurs pretty close to x = -.3. All that's necessary is to simultaneously graph y = 2 and y = 19 x + 8 cos(x), and see where they intersect.
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**** Query problem 6.6.66 19x + 8 cos(x) = 0
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Your solution:
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):
------------------------------------------------
Self-critique Rating:
@&
Expanded solution:
The equation
19 x + 8 cos(x) = 2
is easily rearranged to the form
8 cos(x) = 2 - 19 x.
Each side of the equation is an easily-graphed simple function. If the graphs of the two functions intersect, then at the corresponding value of x, the equation is solved.
A graph of 2 - 19 x has slope -19 and y-intercept 2.
A graph of 8 cos(x) has amplitude 8 and y intercept (0, 8). Its first intercept with the positive x axis is at (pi/2, 0).
It is clear with any reasonable hand-sketched graph that these two graphs do not intersect anywhere to the right of the y axis. The y intercept of the linear graph is lower than that of the cosine graph, and the linear graph is far too steep to intersect the cosine function as it moves to the right. By the time x = pi / 2, the value of y = 2 - 19 x is already less than -20, and the lowest point on the cosine graph is at y = -8.
It is equally clear that the graph of the linear function will intersect the cosine graph to the left of the y axis, and that because of its steepness it will do so rather quickly. By the time the cosine graph reaches the point (-pi/2, 0), the graph of y = 2 - 19 x is greater than 30, far exceeding the peak value 8 of the cosine function. The linear function reaches value 8 at approximately x = -.32, and must therefore have intersected the graph of the cosine function before reaching value 8. A reasonable estimate of the x value of the intersection point would therefore be about -.3.
Using a graphing calculator it is easy to verify that 19 x + 8 cos(x) = 2 occurs pretty close to x = -.3. All that's necessary is to simultaneously graph y = 2 and y = 19 x + 8 cos(x), and see where they intersect.
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#*&!
**** Query problem 6.6.66 19x + 8 cos(x) = 0
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Your solution:
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):
------------------------------------------------
Self-critique Rating:
@&
Expanded solution:
The equation
19 x + 8 cos(x) = 2
is easily rearranged to the form
8 cos(x) = 2 - 19 x.
Each side of the equation is an easily-graphed simple function. If the graphs of the two functions intersect, then at the corresponding value of x, the equation is solved.
A graph of 2 - 19 x has slope -19 and y-intercept 2.
A graph of 8 cos(x) has amplitude 8 and y intercept (0, 8). Its first intercept with the positive x axis is at (pi/2, 0).
It is clear with any reasonable hand-sketched graph that these two graphs do not intersect anywhere to the right of the y axis. The y intercept of the linear graph is lower than that of the cosine graph, and the linear graph is far too steep to intersect the cosine function as it moves to the right. By the time x = pi / 2, the value of y = 2 - 19 x is already less than -20, and the lowest point on the cosine graph is at y = -8.
It is equally clear that the graph of the linear function will intersect the cosine graph to the left of the y axis, and that because of its steepness it will do so rather quickly. By the time the cosine graph reaches the point (-pi/2, 0), the graph of y = 2 - 19 x is greater than 30, far exceeding the peak value 8 of the cosine function. The linear function reaches value 8 at approximately x = -.32, and must therefore have intersected the graph of the cosine function before reaching value 8. A reasonable estimate of the x value of the intersection point would therefore be about -.3.
Using a graphing calculator it is easy to verify that 19 x + 8 cos(x) = 2 occurs pretty close to x = -.3. All that's necessary is to simultaneously graph y = 2 and y = 19 x + 8 cos(x), and see where they intersect.
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#*&!#*&!
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I've inserted a number of notes, especially when the solution given in the original document was either incorrect or could benefit from expansion.
Let me know if you have questions.
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