Inverse Functions

Inverse Sine Function

Range:

On this interval, y=arcsin x.

Evaluating the Inverse Sine Function

                                    

For this problem we would find an angle whose sine is equal to -1/2.





on the interval 




So we get:

Inverse Cosine Function

                                    Range: 

On this interval, y=arccos x. 

For this problem, we need to find an angle whose cosine equals: 

,  so

Inverse Properties

sin(arcsin x )= x  and arcsin(siny) = y when

   and

cos(arccos x) = x and arccos(cosy) = y when 

and 

tan(arctan x) = x and arctan(tan y) = y  when

x is a real number and 

-5 lies on the domain of arctan x, so we can use the inverse property, and we get:

Graphs of Tangent, Cotangent, Cosecant, and Secant functions

Graph of the Tangent Function

The tangent function is odd, therefore the graph y=tanx is symmetric with respect to the origin. 

tan(θ) =sin(θ)/cos(θ)

tangent is then undefined when cosx=0. When there is a zero in the denominator, you put a vertical asymptote there.

The vertical asymptotes will be at –π/2, π/2, and3π/2

Tangent will be zero when the numerator Sinx=0. The x intercepts will then be at 0, π, 2π, - π, -2π

The period for tangent functions is  therefore for y=tanx, the period is  π

Graph of the Cotangent Function

Lets start with y=cotx=Cosx/Sinx

Because Sinx=0 would make the function undefined, values that make it Sinx=0 will be the Vertical Asymptotes.

The vertical asymptotes are at -2π, -π, π, 2π

Because Cosx=0 would make Cotangent 0, values that make Cosx=0 will be the X-Intercepts

The x-intercepts will be –π/2, π/2, and3π/2

y=cotx, the period is  π

Graphs of Secant functions

secx=1/cosx

The asymptotes of secx will be the same of tanx, which are -, , 3π/2, -3π/2

It is important to draw out the reciprocal function and take the reciprocals of the y-coordinate to obtain points for the graph.

The period would be still 2

Because secx is even, it will be centered around the origin.

Graphs of Cosecant functions

cscx=1/sinx

The asymptotes of cscx will be the smae of cotx which are –π/2, π/2, and3π/2

It is important to draw out the reciprocal function and take the reciprocals of the y-coordinate to obtain points for the graph.

Period is still 2π

Because cscx is odd, the graph will be centered around the origin.

Fundamental Identities

If an equation contains one or more variables and is valid for all replacement values of the variables for which both sides of the equation are defined, then the equation is known as an identity.

The Unit Circle

The Unit Circle 

 

What is the Unit Circle?

The unit circle is a circle whose center is at the origin with a radius of one. Because the radius is 1, you can directly measure sine and cosine. If a point on the circle is on the terminal side of an angle in standard position, then the sine of such an angle is the y-coordinate of the point, and the cosine of the angle is the x-coordinate of the point.
 

 

The Pythagorean Theorem states that for a right angled triangle, the square of the long side equals the sum of the squares of the other two sides:

 

x²+ y² = 1²

But 1² is just 1, so:

x²+ y² = 1
(the equation of the unit circle)

Using Coordinates to Find Trigonometric Functions

The coordinates x and y are two functions of the real variable theta. You can use the coordinates to define the six trigonometric functions of theta.

The point of the unit circle is to make math easier and neater. For example, in the unit circle, you have, for any angle theta, the trig values for sine and cosine are sin(θ) = yand cos(θ) = x. Working from this, you can take the fact that the tangent is defined as being tan(θ) = y/x, and then substitute for x and y to easily prove that the value of tan(θ)also must be equal to the ratio sin(θ)/cos(θ).

Special Points of Interest on the Unit Circle

It is very important to memorize the sine and cosine of the angles created by special right triangles for future use.

 

 

Here is a video on how to easily remember the Unit Circle-

For further help, please watch these videos-
http://www.youtube.com/watch?v=ZffZvSH285c
http://www.youtube.com/watch?v=DIGoK51u0KQ
http://www.youtube.com/watch?v=3GgO7Q_kg8Q

 

 -Kenji Johnston

 



 



Questions/Muddy points for Chapter 2

Please comment on this post using the link below.

Write a question or concern you currently have regarding any of the material we've covered so far. 

Rational Functions

For graphing rational functions we need to follow these steps: 

1)  Find any intercepts, if there are any.  (Remember to find the y-intercept with f(0) and x intercepts by setting the numerator equal to zero).

2)  Find the vertical asymptotes by setting the denominator equal to zero and solving.

3)  Find the horizontal asymptote.

4)  The vertical asymptotes will divide the number line into regions.  In each region graph at least one point in each region.  This point will tell us whether the graph will be above or below the horizontal asymptote and if we need to we should get several points to determine the general shape of the graph.

5)   Sketch the graph.

Example:

Sketch the graph of the following function.

                                                            

This time notice that if we were to plug in  into the denominator we would get division by zero.  This means there will not be a y-intercept for this graph.  We have however, managed to find a vertical asymptote already.

Now, let’s see if we’ve got x-intercepts.

                                     

So, we’ve got two of them.

We’ve got one vertical asymptote, but there may be more so let’s go through the process and see.

                           

So, we’ve got two again and the three regions that we’ve got are ,  and .

Next, the largest exponent in both the numerator and denominator is 2 so by the fact there will be a horizontal asymptote at the line,

                                                                 

Now, one of the x-intercepts is in the far left region so we don’t need any points there.  The other x-intercept is in the middle region.  So, we’ll need a point in the far right region and as noted in the previous example we will want to get a couple more points in the middle region to completely determine its behavior.

                                          

Here is the sketch for this function.