#$&* course Mth 164 2/18 9:51 **** Query problem 6.6.12 cot(2`theta/3) = -`sqrt(3) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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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 .........................................23:43:44......!!!!!!!!................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** 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)......!!!!!!!!................................... YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2/3 theta + n * pi = 5 pi / 4 n = 0 2/3 theta + n * pi = 5 pi / 4 2/3 theta + 0 * pi = 5 pi / 4 2/3theta = 5pi/4 theta = 15 pi / 8 n = 1 2/3 theta + n * pi = 5 pi / 4 2/3 theta + 1 * pi = 5 pi / 4 2/3theta + pi = 5pi/4 theta = 27 pi / 8 n = 2 2/3 theta + n * pi = 5 pi / 4 2/3 theta + 2 * pi = 5 pi / 4 2/3 theta + 2 pi = 5 pi / 4 theta = 39 pi / 8 n = 3 2/3 theta + n * pi = 5 pi / 4 2/3 theta + 3 * pi = 5 pi / 4 2/3 theta + 3pi = 5 pi / 4 theta = 51 pi / 8 n = 4 2/3 theta + n * pi = 5 pi / 4 2/3 theta + 4 * pi = 5 pi / 4 theta = 63 pi / 8 n = 5 2/3 theta + n * pi = 5 pi / 4 2/3 theta + 5 * pi = 5 pi / 4 2/3 theta + 5pi = 5 pi / 4 theta = 75 pi / 8 n = 6 2/3 theta + n * pi = 5 pi / 4 2/3 theta + 6* pi = 5 pi / 4 2/3 theta + 6pi = 5 pi / 4 theta = 87 pi / 8 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. .........................................23:57:51......!!!!!!!!................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** How many of these values result in `theta values between 0 and 2 `pi?......!!!!!!!!...................................23:59:12 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is only two value that gives use a value between 0 and 2pi and they are -1 and 0. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given 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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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): theta = (15 + 12 n) pi / 8, for integer values of n 0 <= theta < 2 pi means 0 <= (15 + 12 n ) pi / 8 < 2 pi. 0 <= (15 + 12 n) pi < 16 pi 0 <= 15 + 12 n < 16 so that -15 <= 12 n < 1 and -5/4 < n < 1/12 ------------------------------------------------ Self-critique Rating:2 **** Query problem 6.6.44 solve sin^2(`theta) = 2 cos(`theta) + 2......!!!!!!!!...................................14:41:12 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sin^2(`theta) = 2 cos(`theta) + 2 1 - cos^2(theta) = 2cos(theta) + 2 1 - 2cos(theta) -cos^2(theta) = 0 (1+cos(theta))^2 = 0 1+cos(theta) = 0 cos(theta) = -1 answer 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. **......................................... 14:43:04......!!!!!!!!................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query problem 6.6.66 19x + 8 cos(x) = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 19x + 8 cos(x) = 0 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).......................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): What should I have done for this problem solve for X?
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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).......................................... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): What should I have done for this problem solve for X?