The domain of the tangent function does not include any values of x that are odd multiples of π/2 .

The range of the sine function is from [-1, 1].

The period of the tangent function is π, whereas the period for both sine and cosine is 2π.

The range of a sine wave is altered by the coefficient placed in front of the base equation. So, if you have , this means that the highest point on the wave will be at  and the lowest at .  However, if you then begin to shift the equation vertically by adding values, as in, , then you need to account for said shift.  This would make the minimum value to be  and the maximum value to be . For our question, then, it is fine to use . The  for the function parameter only alters the period of the equation, making its waves "thinner."

The range of a sine wave is altered by the coefficient placed in front of the base equation. So, if you have , this means that the highest point on the wave will be at  and the lowest at ; however, if you then begin to shift the equation vertically by adding values, as in, , then you need to account for said shift. This would make the minimum value to be  and the maximum value to be .  

For our question, the range of values covers . This range is accomplished by having either  or  as your coefficient. ( merely flips the equation over the -axis. The range "spread" remains the same.) We need to make the upper value to be  instead of . To do this, you will need to add  to . This requires an upward shift of .  An example of performing a shift like this is:

Among the possible answers, the one that works is:

The  parameter does not matter, as it only alters the frequency of the function.

The range of the sine and cosine functions are the closed interval from the negative amplitude and the positive amplitude. The amplitude is given by the coefficient,  in the following general equation: 
. Thus we see the range is:

The range of a sine or cosine function spans from the negative amplitude to the positive amplitude. The amplitude is  in the general formula: 
Thus we see amplitude of our function is  and so the range is:

Recall that .

Therefore, we are looking for  or .

Now, this has a reference angle of , but it is in the third quadrant. This means that the value will be negative. The value of  is . However, given the quadrant of our angle, it will be .

You can begin by imagining a little triangle in the fourth quadrant to find your reference angle.  It would look like this:

Now, to find the sine of that angle, you will need to find the hypotenuse of the triangle. Using the Pythagorean Theorem, , where  and  are leg lengths and  is the length of the hypotenuse, the hypotenuse is , or, for our data:

The sine of an angle is:

For our data, this is:

Since this is in the fourth quadrant, it is negative, because sine is negative in that quadrant.

You can begin by imagining a little triangle in the third quadrant to find your reference angle. It would look like this: 

Now, to find the sine of that angle, you will need to find the hypotenuse of the triangle. Using the Pythagorean Theorem, , where  and  are leg lengths and  is the length of the hypotenuse, the hypotenuse is , or, for our data:

The sine of an angle is:

For our data, this is:

Since this is in the third quadrant, it is negative, because sine is negative in that quadrant.

Recall that the  triangle appears as follows in radians:

Now, the sine of  is . However, if you rationalize the denominator, you get:

Since , we know that  must be represent an angle in the third quadrant (where the sine function is negative). Adding its reference angle to , we get:

Therefore, we know that:

, meaning that 

Recall that the sine wave is symmetrical with respect to the origin. Therefore, for any value , the value for  is . Therefore, if  is , then for , it will be .