This one is not as straight-forward.  We must manipulate the original equation before squaring both sides.

                                     (Pythagorean Identity)

                                                                 (divide both sides by 2)

Solving for each:

   radians

Or 

From the unit circle we know that  when .

So now we must go back and check all of our solutions.

Checking 

Checking  (this is also equal to checking )

Both of our solutions are correct.

Our first line of defense when solving trigonometric functions is using a familiar identity/formula such as the Pythagorean Identities or the Double Angle Formulas.  When we are unable to use an identity or formula we are able to square both sides of the equation and with further manipulation we are usually able to produce one of these identities thus simplifying our problem and making it easier to solve.

We can use DeMoivre's formula which states:

Now plugging in our values of  and  we get the desired result.

First, convert this complex number to polar form.

Since the point has a positive real part and a negative imaginary part, it is located in quadrant IV, so the angle is .

This gives us 

To evaluate, use DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get:

 simplifying

,   is coterminal with  since it is an even multiple of 

First convert this point to polar form:

Since this number has a negative imaginary part and a positive real part, it is in quadrant IV, so the angle is 

We are evaluating 

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get:

 which is coterminal with  since it is an odd multiplie

First, convert the complex number to polar form:

Since both the real and the imaginary parts are positive, the angle is in quadrant I, so it is 

This means we're evaluating

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

First, evaluate . We can split this into  which is equivalent to 

[We can re-write the middle exponent since  is equivalent to ]

This comes to 

Evaluating sine and cosine at  is equivalent to evaluating them at  since 

This means our expression can be written as: 

Begin by converting the complex number to polar form:

Next, put this in its generalized form, using k which is any integer, including zero:

Using De Moivre's theorem, a fifth root of  is given by:

Assigning the values  will allow us to find the following roots. In general, use the values .

These are the fifth roots of .

Begin by converting the complex number to polar form:

Next, put this in its generalized form, using k which is any integer, including zero:

Using De Moivre's theorem, a fifth root of 1 is given by:

Assigning the values  will allow us to find the following roots. In general, use the values .

These are the cube roots of 1.

Begin by converting the complex number to polar form:

Next, put this in its generalized form, using k which is any integer, including zero:

Using De Moivre's theorem, a fifth root of  is given by:

Assigning the values  will allow us to find the following roots. In general, use the values .

These are the fifth roots of .

Given the polar coordinates , the  -coordinate is  . We can find this coordinate by substituting :

Likewise, given the polar coordinates , the  -coordinate is  .  We can find this coordinate by substituting :

Therefore the rectangular coordinates of the point  are .