course MTH 164
......!!!!!!!!...................................
14:42:45 `q001. Construct the triangle corresponding to the pi/6 angular position: Sketch a segment from the origin to the pi/6 position on the circle, then straight down to the x-axis and back along the x-axis to the origin. Sketch also the triangle corresponding to the theta = -pi/6 position, sketching a segment from the origin to the -pi/6 position on the circle, then straight up to the x-axis and back along the x-axis to the origin. What are the three angles in each of the two triangles you have constructed? The two triangles you have sketched join to form a larger triangle. Describe the larger triangle. How do the lengths of the sides of this larger triangle compare?
......!!!!!!!!...................................
RESPONSE --> 30, 60, 90 A right triangle The larger triangle all angles are 60 the length of the side parallel to the y-axis is twice that of the seperate triangles.
.................................................
......!!!!!!!!...................................
14:47:08 Figure 2 shows the construction of these triangles on a unit circle, whose radius is 1. The angle made by the vertical and the horizontal sides of each triangle is a right angle of pi/2 radians. The remaining two angles must add up to a right angle, and since the central angle of each triangle is pi/6 the other angle is pi/3. These two triangles naturally join to form a larger triangle, which is easily shown to be an equilateral triangle: The central angle of the larger triangle is pi/6 + pi/6 = pi/3. The other two angles are also pi/3, so the triangle is equiangular. If the triangle is rotated through an angle of pi/3 it coincides with itself, and this symmetry implies that the triangle is equilateral. Therefore the three sides of the larger triangle are equal.
......!!!!!!!!...................................
RESPONSE --> check though figure 2 isn't what you claim it to be. Perhaps you were refering to figure 1.
.................................................
......!!!!!!!!...................................
14:48:52 `q002. What therefore is the length of the vertical side of each of the two original triangles?
......!!!!!!!!...................................
RESPONSE --> Individualy, Sin(pi/6) = .5 Together, 2 Sin(pi/6) = 1
.................................................
......!!!!!!!!...................................
14:50:08 The vertical side of the equilateral triangle is formed from the vertical sides of the two original triangles. Since the original two triangles are congruent it follows that the vertical side of each is 1/2 the side of the equilateral triangle. Since every side of the equilateral triangle is 1, it follows that the vertical sides of the original triangles are each 1/2.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
14:55:16 `q003. Each of the original triangles has hypotenuse 1 and vertical leg 1/2. By the Pythagorean Theorem, what therefore is the length of the horizontal side of either triangle?
......!!!!!!!!...................................
RESPONSE --> 1^2 = S^2 + .5^2 1 - .25 = S^2 .75 = S^2 0.87
.................................................
......!!!!!!!!...................................
14:58:20 By the Pythagorean Theorem c^2 = a^2 + b^2, where a and b are the legs and c the hypotenuse of a right triangle. Thus b^2 = c^2 - a^2 so that b = sqrt(c^2 - a^2). In the case of this triangle, if a = 1/2 and c = 1 we get b = sqrt(1^2 - (1/2)^2) = sqrt(1 - 1/4) = sqrt(3/4)= sqrt(3) / 2.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
15:00:03 `q004. What therefore is the y coordinate of the point on the unit circle corresponding to angular position pi/6? What is the x coordinate?
......!!!!!!!!...................................
RESPONSE --> (sqrt(3)/2, .5)
.................................................
......!!!!!!!!...................................
15:00:33 The triangle is depicted in Figure 65. The y coordinate is 1/2 and the s coordinate is sqrt(3) / 2. The unit-circle point corresponding to angular position pi/6 is (sqrt(3)/2, 1/2), as indicated in Figure &&. Note that the approximate coordinates, accurate to 2 significant figures, are the values we used previously for this angle so that the approximate coordinates are (.87, .5).
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
15:04:33 `q005. Construct the angle corresponding to the angular position pi/4. Starting at the origin draw a line out to the pi/4 position, then directly down to the x axis and finally back along the x axis to the origin; this will form a triangle. Draw another triangle, this time starting from the origin, go into the pi/4 point, then straight back to the y-axis along horizontal line, and finally down the y-axis to the origin. What are the three angles in each triangle? How do the lengths of the two legs of each triangle compare? When these two triangles are put together what shape do they form?
......!!!!!!!!...................................
RESPONSE --> 45, 45, 90 They are the same length. Together you have a square.
.................................................
......!!!!!!!!...................................
15:06:06 The triangles are shown in Figure 46. Each triangle has a horizontal and a vertical side, so one of the angles formed is a right angle pi/2. The central angle in each is pi/4, so it follows that the remaining angle, in order that the three angles add up to pi, is also pi/4. These triangles, having two equal angles, are therefore both isosceles, with equal legs. The two triangles form a square, since the sides of the combined figure are all horizontal or vertical and since the equal legs form the sides. If the entire figure is flipped about the theta = pi/4 line there will be no change except that the x and y sides will be interchanged. The interchanged triangles will coincide, but the x and y sides of the triangles will be interchanged, showing that the x and y coordinates of the two triangles are the same.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
15:08:08 `q006. If the circle has radius 1 it follows that the hypotenuse of either triangle is 1. Let s stand for the lengths of the (equal) legs of either triangle. By the Pythagorean Theorem what must therefore be the length of the side s?
......!!!!!!!!...................................
RESPONSE --> c^2 = a^2 + b^2 a^2 and b^2 are the same value. c^2 = 2s^2 1/2 = s^2 .71
.................................................
......!!!!!!!!...................................
15:11:14 Figure 23 shows the first triangle with the legs labeled s and the hypotenuse 1. The Pythagorean Theorem tells us that s^2 + s^2 = 1, so that 2 s^2 = 1 and s^2 = 1/2. From this we get s = 1 / sqrt(2) = sqrt(2) / 2.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
15:11:53 `q007. What therefore is the y coordinate of the point corresponding to angular position pi/4? What is the x coordinate?
......!!!!!!!!...................................
RESPONSE --> (.71, .71)
.................................................
......!!!!!!!!...................................
15:12:06 The y coordinate is the same as the x coordinate, sqrt(2) / 2. The unit-circle point corresponding to angular position pi/4 is therefore (sqrt(2) / 2, sqrt(2) / 2), as indicated in Figure 83. Note that the approximate coordinates, accurate to 2 significant figures, are the values we used previously for this angle so that the approximate coordinates are (.71, .71).
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
15:17:20 `q008. Sketch the triangle corresponding to the angular position pi/3. Explain why this triangle is similar to the triangle you constructed for angular position pi/6. What therefore are the x and y coordinates of the theta = pi/3 point on the unit circle?
......!!!!!!!!...................................
RESPONSE --> this creates the same triangle with a different orientation. The angle of 60 lies closer to the origin now. (.5, .87)
.................................................
......!!!!!!!!...................................
15:17:34 The triangle is shown in Figure 88. The horizontal and vertical sides are reversed from the pi/6 triangle. So the coordinates of the points are also reversed to give us (1/2, sqrt(3)/2).
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
15:32:55 `q009. Give the x and y coordinates of the points on the unit circle corresponding to angles which are multiples of pi/6. Sketch the unit circle and all angles which are multiples of pi/6, in give their coordinates.
......!!!!!!!!...................................
RESPONSE --> 0, pi/6, pi/3, pi/2, 2pi/3, 5pi/6, pi, 7pi/6, 4pi/3, 3pi/2, 5pi/3, 11pi/6 the easiest way to calculate these, (cos(x), sin(x)) 1,0 87,.5 .5,.87 0,1 -.5,.87 -.87,.5 -1,0 -.87,-.5 -.5,-.87 0,-1 .5,-.87 87,-.5 1,0
.................................................
......!!!!!!!!...................................
15:33:58 Figure 50 shows the angles 0, pi/6, pi/3, pi/2, 3 pi/2, 5 pi/6, ..., 2 pi with accurate values for the coordinates of the points.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
15:40:32 `q010. Give the x and y coordinates of the points on the unit circle corresponding to angles which are multiples of pi/4. Sketch the unit circle and all angles which are multiples of pi/6, in give their coordinates.
......!!!!!!!!...................................
RESPONSE --> 0 1,0 pi/4 sqrt(2)/2,sqrt(2)/2 pi/2 0,1 3pi/4 -sqrt(2)/2,sqrt(2)/2 pi -1,0 5pi/4 -sqrt(2)/2,-sqrt(2)/2 3pi/2 0,-1 7pi/4 sqrt(2)/2,-sqrt(2)/2
.................................................
......!!!!!!!!...................................
15:41:05 Figure 28 shows the angles 0, pi/4, pi/2, 3 pi/4, 2 pi, 5 pi/4, ..., 2 pi with accurate values for the coordinates of the points.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
15:48:26 `q011. Make the table for sin(theta) vs. theta with 0 <= theta < 2 pi, using an increment of pi/4. Sketch the corresponding graph. Now reverse the columns of the table and sketch the associated graph. Does your first graph represent a function? Does your second graph represent a function?
......!!!!!!!!...................................
RESPONSE --> sin(x) vs x first graph sin(x) second graph sin-1(x)
.................................................
......!!!!!!!!...................................
15:49:42 Using .71 to stand for sqrt(2) / 2 (your table should use sqrt(2)/2 where this one uses .71) we have the following table: theta sin(theta) 0 0.0 pi/4 0.71 pi/2 1.0 3 pi/4 0.71 pi 0.0 5 pi/4 -0.71 3 pi/2 -1.0 7 pi/4 -0.71 2 pi 0.0. The reversed table is sin(theta) theta 0.0 0 0.71 pi/4 1.0 pi/2 0.71 3 pi/4 0.0 pi -0.71 5 pi/4 -1.0 3 pi/2 -0.71 7 pi/4 0.0 2 pi. The corresponding graphs are indicated in figure 68. The red dots indicate points on the table; the rest of the function is filled in from what we know about the shape of the graph of y = sin(x). The graph of the sine function is the usual graph. The graph of the reversed column doesn't represent a function because for example the horizontal coordinate zero is associated with two different values, 0 and pi (and also 2 pi). Another way of saying this is that the function fails the vertical line test, which says that a graph can represent a function only if any vertical line intersecting the graph of the function intersects the graph in only 1 point. Still another way of saying the same thing is that any function must be single-valued, with any number you put into the function giving you exactly 1 value. For most numbers the first column of the inverted table we get 2 values. The numbers in the first column usually appear twice. For example sqrt(2)/2 in the first column is listed next to pi/4 and also next to 3 pi/4, 0 is listed with 0 and also with pi, etc..
......!!!!!!!!...................................
RESPONSE --> oh, the vertical line test. My mistake.
.................................................
......!!!!!!!!...................................
15:52:53 `q012. Now restrict the original table of sin(theta) vs. theta to values of theta for which pi/2 <= theta <= 3 pi/2, then reverse columns of the table. Does the resulting table give you a function? Sketch the graph corresponding to the table.
......!!!!!!!!...................................
RESPONSE --> yes, there is still a function.
.................................................
......!!!!!!!!...................................
15:54:21 The table is as follows theta sin(theta) pi/2 1.0 3 pi/4 0.71 pi 0.0 5 pi/4 -0.71 3 pi/2 -1.0 The inverted table is as follows: sin(theta) theta 1.0 pi/2 0.71 3 pi/4 1.0 pi -0.71 5 pi/4 -1.0 3 pi/2 Each number in the first column appears exactly once in the table for the inverted function. The corresponding graphs are shown in figure 14. We see that the sine-function graph passes a horizontal line test, so that the inverted function passes a vertical-line test. Any vertical line passing through the inverted graph passes through at exactly point, showing that the graph does define a function.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
15:57:25 `q013. If we make a table of sin(theta) vs. theta for -pi/2 <= theta <= pi/2, can this table be inverted to give us a function?
......!!!!!!!!...................................
RESPONSE --> -pi/2 to pi/2 yes, this graph can be inverted to give a function.
.................................................
......!!!!!!!!...................................
15:57:38 The inverted table is as follows: The table for the sine function is as follows theta sin(theta) -pi/2 0 -pi/4 -0.71 0 0 pi/4 0.71 pi/2 1.0 The inverted table is as follows: sin(theta) theta -1.0 -pi/2 -.71 -pi/4 0.0 0 0.71 pi/4 1.0 pi/2 Each number in the first column appears exactly once. The corresponding graph is shown in figure 34. Any vertical line passing through the graph passes through at exactly point, showing that the graph does define a function.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
15:59:56 `q014. Either of the functions graphed in the previous two exercises constitute inverses of portions of the sine function. Re-label the last table you made (the inverted table), labeling the first column x and the second column arcsin(x). This function is the generally accepted inverse of the sine function. What are the values of arcsine (.71), arcsine (0) and arcsin(-1)?
......!!!!!!!!...................................
RESPONSE --> arcsine (.71) = pi/4 arcsine (0) = 0 arcsine (-1) = 3pi/2
.................................................
......!!!!!!!!...................................
16:02:21 The table is as follows: sin(theta) theta -1.0 -pi/2 -.71 -pi/4 0.0 0 0.71 pi/4 1.0 pi/2 Arcsine(.71) is the number in the second column of the table across from .71 in the first column. This number is pi/4. Arcsine(0) is the number in the second column of the table across from 0 in the first column. This number is 0. Arcsine(-1) is the number in the second column of the table across from 1 in the first column. This number is -pi/2.
......!!!!!!!!...................................
RESPONSE --> check.
.................................................
......!!!!!!!!...................................
16:03:48 `q015. What are the solutions to the following equations: arcsin(x) = pi/4; arcsin(x) = -pi/3; sin(x) = 1.
......!!!!!!!!...................................
RESPONSE --> x = .71 x = -.5 x = 0
.................................................
......!!!!!!!!...................................
16:04:07 If arcsin(x) = pi/4 then x appears next to pi/4 on the table of sin(x) vs. x. In other words sin(pi/4) = x. Since sin(pi/4) = sqrt(2)/2 we have x = sqrt(2)/2. }{Another way of putting this is to say arcsin(sqrt(2)/2) = pi/4. If arcsin(x) = -pi/3, similarly, this means that x = sin(-pi/3) so that x = -1/2.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
16:12:40 `q016. Find solutions to the following equations: sin(2x) = .87; sin(3x/4 + pi) = -.5?
......!!!!!!!!...................................
RESPONSE --> sin(2x) = .87 2x = pi/3 x = pi/6 (I think I made an error on the previous question) sin(3x/4 + pi) = -.5 3x/4 + pi = pi/6 3x/4 = -5pi/6 9x/12 = -10pi/12 9x = -10pi x = -1.11pi
.................................................
......!!!!!!!!...................................
16:17:23 Starting with sin(2x) = .87, apply the arcsin function to both sides of the equation. We get the equation arcsin ( sin(2x) ) = arcsin(.87). The arcsin and the sine function are inverse functions; one 'undoes' the other so that arcsin(sin(2x)) = 2x. So we obtain 2x = arcsin(.87). We know from the table of the sine function that arcsin(.87) = pi/3 so we have 2x = pi/3. We divide both sides by 2 to obtain x = pi/6. Using a similar strategy for the second equation we write arcsin( sin(3x/4) ) = arcsin(-.5), which gives us 3x/4 = -pi/6 so that x = 4/3(-pi/6) = -2 pi / 9.
......!!!!!!!!...................................
RESPONSE --> sin(3x/4 + pi) = -.5 was the second problem.
.................................................
......!!!!!!!!...................................
17:00:54 `q017. Find solutions to the following equations: sin(3x) = .38; sin(2x+pi/4) = .93.
......!!!!!!!!...................................
RESPONSE --> 3x = sin-1(.38) 3x=22.33 x=7.44 2x+pi/4 = sin-1(.93) 2x+pi/4 = 68.43 2x = 68.43 - pi/4 x = 33.82
.................................................
......!!!!!!!!...................................
17:01:24 If sin(3x) = .38 then we have arcsin( sin(3x) ) = arcsin(.38). .38 doesn't appear on any of our tables so we have to find arcsin(.38) using a calculator. The arcsin key is sometimes labelled arcsin, sometimes sin^-1, and is often the 2d function on the sine key. We find that arcsin(.38) is about .39. So we have 3x = .39, so that []x = .39 / 3 = .13. If sin(2x+pi/4) = .93 then arcsin(2x + pi/4) = arcsin(.93). Using the calculator to find arcsin(.93) we get 2x + pi/4 = 1.19, approx.. We easily solve for x to obtain x = (1.19 - pi/4) / 2 = .2 approx.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
17:05:09 `q018. For what values of t do we have sin(t + pi/3) equal to .5? Give the values for which t + pi/3 lies between 0 and 2 pi. Are there values of t for which t + pi/3 lies outside the interval from 0 to 2 pi?
......!!!!!!!!...................................
RESPONSE --> t = -pi/6 X t = pi/2 Check t without the limiting X could be an infinite amount of values. Not only are the negatives considered, but complete cycles are as well.
.................................................
......!!!!!!!!...................................
17:12:24 sin(t+pi/3) = .5 means that sin(theta) = .5, with theta = t + pi/3. The reference-circle picture of sin(theta) = .5 is given in Figure 60. We see that this occurs at the angle arcsin(.5) = pi/6 as well as at 5 pi/6. So sin(t + pi/3) = .5 for t + pi/3 = pi/6 and for t + pi/3 = 5 pi/6. We can solve each of these equations for t: t + pi/3 = pi/6 is solved by adding -pi/3 to both sides to obtain t = - pi/6. t + pi/3 = 5 pi/6 is similarly solved to get t = pi. Any angle which is coterminal with either theta = pi/6 or with 5 pi/6 (e.g., pi/6 + 2 pi, pi /6 + 4 pi, pi/6 + 6 pi, etc. or 5 pi/6 + 2 pi, 5 pi/6 + 4 pi, etc. ) will yield the same y coordinate and therefore the same value of the sine.
......!!!!!!!!...................................
RESPONSE --> 0 to 2pi, okay... on a side note, did you mean to say t = pi/2 not t = pi?
.................................................
......!!!!!!!!...................................
17:19:30 `q019. What are the first four positive values of t for which sin(t + pi/3) = .5?
......!!!!!!!!...................................
RESPONSE --> sin-1(.5) = pi/6 t + pi/3 = pi/6 pi/2 + pi/3 = pi/6 11pi/6 + pi/3 = 13pi/6 5pi/2 + pi/6 = 17pi/6 21pi/6 + pi/3 = 23pi/6
.................................................
......!!!!!!!!...................................
17:21:33 As seen in the preceding problem sin(t + pi/3) = .5 when t + pi/3 = pi/6, 5 pi/6, pi/6 + 2 pi, 5 pi/6 + 2 pi, pi/6 + 4 pi 5 pi/6 + 4 pi, etc.. We also saw that the first two solutions, corresponding to theta = pi/6 and theta = 5 pi/6, are t = -pi/6 and t = pi. Only the second of these solutions is positive. The solutions corresponding to theta = t + pi/3 = pi/6 + 2 pi, 5 pi/6 + 2 pi, pi/6 + 4 pi, 5 pi/6 + 4 pi, etc., will also be positive. Setting t + pi/3 equal to each of these angles and solving for t 2e obtain solutions t = 11 pi / 6, t = 5 pi/2, t = 25 pi/6, t = 9 pi/2. The first four positive solutions are t = pi/2, 11 pi/6, 5 pi/2, 25 pi/6.
......!!!!!!!!...................................
RESPONSE --> my solutions might have been wrong. Setting x+pi/3 = sin-1 of .5 was my process however.
.................................................
......!!!!!!!!...................................
17:21:55