The period of the sine function is .  The period of our new function is 1.  Each period will have two zeroes and there will be one tacked on at the end when the domain is closed.  There will be -intercepts at . 

The graph is below.

It is a question of what quadrant is in.

A negative value for secant indicates quadrant II or III. Since secant is the reciprocal of cosine, the measurement includes the x value and the r value with regards to position.To get a negative value for secant or cosine we will need a negative x value and either a positive or negative y value to get the correct r value.

A positive value for cotangent indicates quadrant I or III. Since cotangent is the reciprocal of tangent, the measurement includes the x and y values with regards to the position. To get a cotangent that is positive we will need a positive x value and either a positive or negative y value.

The overlap between these two statements is quadrant III. Therefore, must be in quadrant III.

:

There is a restriction for the range of the inverse tangent function from .

The inverse tangent of a value asks for the angle where the coordinate  lies on the unit circle under the condition that .  For this to be valid on the unit circle, the  must be very close to 1, with an  value also very close to zero, but cannot equal to zero since  would be undefined.  

The point  is located on the unit circle when  , but  is invalid due to the existent asymptote at this angle.

An example of a point very close to  that will yield  can be written as:

Therefore, the approximated rounded value of  is .

To determine the value of , solve each of the terms first.

The inverse cosine has a domain and range restriction.

The domain exists from , and the range from .  The inverse cosine asks for the angle when the x-value of the existing coordinate is .  The only possibility is  since the coordinate can only exist in the first quadrant.

The inverse sine also has a domain and range restriction.

The domain exists from , and the range from .  The inverse sine asks for the angle when the y-value of the existing coordinate is .  The only possibility is  since the coordinate can only exist in the first quadrant.

Therefore:

For this particular problem we need to recall that the inverse cosine cancels out the cosine therefore,

.

So the expression just becomes 

From here, recall the unit circle for specific angles such as .

Thus,

.

Rewrite and evaluate .

The inverse sine of one-half is  since  is the y-value of the coordinate when the angle is .

To convert from radians to degrees, replace  with 180.

This one seems complicated, but becomes considerably easier once you implement the fact that the composite  cancels out to 1 and you are left with  which is equal to 1

This one is rather simple with knowledge of the unit circle: the value is extremely close to zero, of which  always

Given the value of the opposite and hypotenuse sides from the sine expression (3 and 4 respectively) we can use the Pythagorean Theorem to find the 3rd side (we’ll call it “t”): . From here we can easily deduce the value of  (the adjacent side over the opposite side)