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It is not clear that you understand and are able to apply the details of the transformation process, though everything you say is true. You simply don't go into enough detail. Be sure you understand all the details in the solutions given below:

** Here are two solutions provided by students from previous years:

The amplitude is 3

The Period = T= (2`pi)/`omega = 2`pi/(2) = `pi

The phase shift = `phi/(`omega) = `pi/2

The graph of y = 3 cos (2x + `pi) will lie between -3 and 3 on the y axis.

One cycle will begin at x = `phi/(`omega) = `pi/2 and will end at 2`pi/(`omega) + `phi/(`omega) = (2`pi)/2 + (`pi)/2 = (3`pi)/2.

We then divide the interval of [`pi/2, 3`pi/2] into (4) subintervals each of length `pi divided by 4 = `pi/4:

[`pi/2, 3`pi/4], [3`pi/4, `pi], [`pi, 5`pi/4], [5`pi/4, 3`pi/2].

The five key points for the graph are:

(`pi/2, 3), (3`pi/4, 0), (`pi, -3), (5`pi/4, 0), and (3`pi/2, 3).

ANOTHER STUDENT SOLUTION (consistent with preceding but with different details provided):

the graph of this function has a maximum point of y=+3 and a minimum point of y=-3. at the origin the graph touches the point y=-3. and whenever x= pi, 2pi and 3pi y=-3. and at the points x= (-pi),-2pi,-3pi y= -3. when x=pi/2 and 3pi/2 and -pi/2 and -3pi/2 y= 3.

to solve for the amplitude and the period and the phase shift we use the equation y= Acos((omega)(x)-phi). so the amplitude of the equation is the absolute value of A which is 3. so A=3.

the period is 2pi/2 which is pi so there is a period at pi.

and the phase shift is phi/omega. which in this case is pi/2. so the phase shift is pi/2.

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Again you need to describe the details of the transformation process:

** Starting with y = cos(x), which has amplitude 1 and period 2 `pi and which peaks on the y axis (i.e., at x = 0) and at every interval of 2 pi on the x axis, we apply the appropriate transformations as follows:

We first multiply the function by 2, which doubles all the y coordinates, stretching the graph vertically by factor 2. This doubles the amplitude from 1 to 2.

Next we multiply x by 2 `pi, which compresses the graph in the horizontal direction by factor 2 `pi. So the period of the function is changed from 2 `pi to 2 `pi / (2 `pi) = 1.

We then replace x by x - 4, which shifts the graph 4 units to the right.

Our graph now has a peak at x = 4. It oscillates between max value y = +2 and min value y = -2, peaking at x = 4 and at regular intervals of 1 so that peaks occur at x = 4, 5, 6, . . . as well as 3, 2, 1, . . . . **

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When you solve the problem you should have noted for you had solve it in your homework and should have transcribed the outline of your solution here.

** 1 -( sin^2(x)/(1-cos x) =

1- (1-cos^2(x))/(1-cos x) =

1- [(1-cos x)(1+cos x)/(1-cos x) =

1- (1+cos x) =

1 - 1 - cos x = - cos x. **

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** There are many ways to rearrange this equation to prove the identity. Here we will start by changing everything to sines and cosines using sec(x) = 1 / cos(x). We get

[ 1 / cos(x) ] / ( 1 + 1 / cos(x) ] = (1 - cos(x) ) / sin^2(x).

Multiplying both sides by the common denominator (1 + 1 / cos(x) ) * sin^2(x) we get

[ 1 / cos(x) ] * sin^2(x) = (1 - cos(x) ) ( 1 + 1 / cos(x) ). Multiplying out the right-hand side and simplifying the left we have

sin^2(x) / cos(x) = 1 + (1 / cos(x)) - cos(x) - 1 or since 1 - 1 = 0 just

sin^2 / cos(x) = [ 1 / cos(x)] - cos(x).

Multiplying both sides by the only remaining denominator cos(x) we have

sin^2(x) = 1 - cos^2(x), which we rearrange into the basic Pythagorean identity

sin^2 x + cos^2 x = 1. **

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I've inserted several notes and solutions. You should be sure you understand every detail of the given solutions; if not, insert self-critiques, questions, etc. into a copy of this document, mark your insertions before and after with &&&&, and submit a copy.