The blue graph in the figure below depicts graph of the y coordinate of motion around a circle of radius 4, starting at the `pi/2 position on the circle. It should be clear that the y coordinate begins at y = 4 and that a complete cycle requires the eight seconds indicated on the graph. Since we are graphing the y coordinate it should be immediately clear that we will model this motion with a sine function.
In order to get the equation for the function whose graph is represented, we begin by adjusting the period of the function. The function y = sin(2 `pi t) completes a period in 1 second, with the angle 2 `pi t going from 0 to 2`pi as t goes from 0 to 1. We adjust this so that as t goes from 0 to 8, the angle goes from 0 to 2 `pi. We accomplish this by making the angle 2 `pi / 8, so that the function becomes y = sin(2 `pi / 8). This function is sketched in read on the graph. We see that it has the appropriate period, but its amplitude is not great enough and it isn't in phase with the desired function.
In order to adjust the amplitude we need only multiply the function by 4.
In order to have the angle equal to `pi/2 at the start, when t = 0, we can simply add `pi / 2 to the angle. The angle therefore becomes 2 `pi / 8 + `pi / 2 (note that 2 `pi / 8 reduces to `pi / 4; we will continue in this example writing 2 `pi / 8 to emphasize that the period is 8; but in reporting our final result we would ordinarily reduce the fraction).
The resulting period is shown in purple at the lower right of the figure below.
We also wish to graph motion starting at 3 `pi / 2, the and requiring three seconds for a complete revolution. The amplitude is still 4 so we leave the 4 in front of the sine function. We adjust the period by making our angular displacement from the original point 2 `pi / 3, so that in 3 seconds the angle will change by 2 `pi. We add 3 `pi / 2 to this angular displacement so that our initial t = 0 angular position is 3 `pi / 2.
We note that graph of y = 4 sin(2 `pi / 8 * t + `pi / 2) is shifted left by 1/4 cycle, corresponding to a horizontal shift of -`pi / 2 radians in the reference angle (the angular position on the circle corresponding to the position on the graph), or by -2 seconds.
We note that graph of y = 4 sin(2 `pi / 8 * t + `pi / 2) is shifted left by 1/4 cycle, corresponding to a horizontal shift of -`pi / 2 radians in the reference angle (the angular position on the circle corresponding to the position on the graph), or by -2 seconds.
We does see that the addition of an angle to the term 2 `pi / 8 * t shifts the angular position to the left by the added angle. In general graph of y = A sin(Bt + c) will have its angular position shifted c units to the left to the graph of the function y = A sin(Bt).
If we factor the 2 `pi / 8 out of the expression 2 `pi / 8 * t + `pi / 2, we obtain 2 `pi / 8 ( t + 2). We see from the example that the graph of y = 4 sin(2`pi / 8 * (t + 2)) is shifted along the t axis by -2 units. Generalize and we will say that y = A sin(B(t + h)) is shifted -h units along the t axis from the function y = A sin(Bt).
The figure below shows how we can quickly create a table for the values of trigonometric functions. The angle `theta is shown in this figure from 0 to `pi, changing by an increment of `pi / 4. The corresponding angular positions on the units circle, and the corresponding x and y coordinates expressed as decimal approximations are shown to the right of the table. In each case the number .71 represents an approximation of the quantity `sqrt(2) / 2, which as you should note is the exact value of either the sine or the cosine of `pi / 4. You should be able to construct such a picture and a table instantly whenever you need it, and should see clearly how to extend the table to `theta values of 2 `pi or more.
The values of the sine and the cosine are indicated in appropriate columns. To obtain the value of the tangent of each angle we note that attention of angle is y / x for the corresponding x and y coordinates of the point on the units circle corresponding to the angle. We will thus divide the value of the sine by the value of the cosine for each angle. This works fine for the first two angles but when we reach `pi / 2 we see that we have an undefined division. If we visualize moving along the unit circle, approaching the `pi /2 position from the first quadrant, we see that the value of the sine approaches 1 while the value of the cosine approaches 0, so that the value of the cosine divides into the value of the sine more and more times. Since there is no limit to how small value of the cosine can become, there is no limit how large the values of the quotient can become. The values of the tangent function therefore approach infinity as the angular position approaches `pi / 2 from first-quadrant angles.
Similarly if we approach `pi / 2 from second-quadrant angular positions, the sine
approaches 1 in the cosine approaches 0. The only differences that the cosine approaches 0
through negative values, so that the quotient will always been negative and the tangent
will approach negative infinity.
The corresponding graph of the tangent function is depicted at the bottom right-hand corner, with the main branch of the function having a positive vertical asymptote at `pi / 2 and in negative vertical asymptote at - `pi / 2.
The inverse sine function is obtained by first restricting our attention to the sine function between ` pi / 2 and `pi / 2, a domain over which the sine function takes all of its possible values, from -1 to 1, taking each value exactly once so that for each y value there is one and only one x value. We can then construct the inverse function, mentally switching the two columns of a table of values for y = sin(x). This process corresponds to reflecting the points of the graph of y = sin(x) through the line y = x, as depicted below.
The figure below shows the unit-circle relationship between a number y between -1 and 1 on the y axis and sin^-1(y). The inverse sine is the angle between -`pi / 2 and `pi / 2 whose sine, or y coordinate on a unit circle, is y.
In an analogous way the inverse cosine of a number x between -1 and 1 is the angle between 0 and `pi corresponding to x coordinate x on the unit circle.
These pictures can be very valuable in thinking about inverse signs and cosines.
A question was raised regarding the text problem which asked for the value of sin(11`pi / 12), given that sin(`pi / 12) = .259. We solve this by constructing a unit circle showing the angle `pi / 12 and its corresponding y coordinate .259. We then locate 11 `pi / 12, which is `pi / 12 short of 12 `pi / 12, or `pi. It is clear by constructing this angle that its y coordinate will be the same is that of `pi / 12.
We can begin to understand the behavior of polynomials by looking at the polynomial f(x) = (x - 3)(x + 2)(x - 5). This is an polynomial since its expanded form will be x^3 + b x^2 + c x + d, which is a cubic polynomial.
The first thing we observe about this polynomial is that it has zeros at x = 3, -2 and 5, since each of these values makes a factor equal to 0. We observe also that no other value of x can make the polynomial zero, since no other value to make one of the factors zero.
We also note that when x is a very large negative number, f(x) will be a really really large negative number, since each of the three factors will be large negative number. Similarly when x is a large positive number each factor will be a large positive number and f(x) will be a really really large positive number.
When all this information is put together we see that the graph of this function must have the overall behavior of the function graphed in purple below. Note that the graph has two 'turning points', where its behavior changes from increasing to decreasing or vice versa. Note also that if the graph had more than two such points, it might be possible to sketch such a graph that passed through the x axis more than three times, which would not be possible for a function like the one we are considering (the cubic polynomial with its 3 linear factors).
Now note that if we added a constant number to f(x), obtaining for example a function like g(x) = (x-3)(x+2)(x-5) + 20, the effect would be to raise the graph vertically while maintaining its shape. If we add a large enough number, we will raise the graph high enough that it only insects the x axis in one point, as is the case with the red graph. So it is possible for a cubic polynomials to have just one zero. This corresponds to a polynomial which factors into a linear factor and an irreducible quadratic factor (that is, a quadratic which has no zeros and hence cannot be factored).
If we add just the right constant number to f(x), we can even get a graph like the green graph below, which intersects the x axis in exactly two points. This can happen in the case where two of the linear factors are identical.
"
