We first need to think about the graph of the function .

Using the formula  where  is the vertical shift, we have to perform a transformation of moving the function  up two units on the graph.

First, we must solve for the inverse of 

So now we are trying to find the range of and plot the function .  Let’s start with the graph of .  We know the domain is .

Now using the formula  where  = Period, the period of   is .  And so we perform a transformation to the graph of  to change the period from  to .

We can see that the graph has a range of  

This is true because just as the range of  is all real numbers due to the vertical asymptotes of the function, the function  extends to all values of  but is limited in its values of  .  No matter the transformations applied, all values of will still be reached.

First, we must consider the graph of .

Using the formula  we can apply the transformations step-by-step.  First we will transform the amplitude, so  so we must shorten the amplitude to .

Now we must apply a vertical shift of one unit since .  This leaves us with our answer. 

This is not true.  This is because when squaring both sides and then plugging back into the original equations, some of our solutions may be extraneous solutions.  Therefore, when solving a trigonometric equation by squaring both sides, all solutions found must be plugged back into the original equation and validated.

We should only square by both sides when all other identities are not able to be used in an equation.  Quite often, you will find that a trigonometric identity can be used to simplify an equation.  Squaring both sides is ultimately trying to produce a trigonometric identity in order to solve for the equation.

We begin with our original equation:

                 (Pythagorean Identity)

Looking at the unit circle we see that  at  and .  We must plug these back into our original equation to validate them.

Checking 

Checking 

And so our only solution is 

We begin with our original equation:

               (Pythagorean Identity)

From the unit circle, we see that . We must check both of these solutions in the original equation.

Checking 

Checking  

So we see our only solution is 

We begin with our original equation

                                  (Pythagorean Identity)

                                                                                            (substitution)

Using this form, we see we really only need to consider when   at   and .  Now we must plug these values into the original equation to check and see if they are both acceptable solutions to our problem.

Checking :

Checking 

By checking our solutions we see the only solution to our equation is .

We begin with our original equation.

                                             (Pythagorean Identity)

                                                                    (Double-Angle Formula)

We know that  will be equal to  for when  is any multiple of and when .  We need to check both solutions (we will simply check  for simplicity) to make sure they are valid solutions.

Checking :

Checking 

By checking our solutions, we see the only solution to this equation is