The range of the function represents the spread of possible answers you can get for , given all values of . In this case, the ordinary range for a cosine function is , since the largest value that cosine can solve to is  (for a cosine of  or a multiple of one of those values), and the smallest value cosine can solve to is  (for a cosine of  or a multiple of one of those values).

However, in this case our final answer is first multiplied by , then decreased by  after the cosine is applied to . Multiplying the initial  range by  results in a new range of . Next, subtracting  from this range gives us a new range of .

Note that the  does not change our range. This is because, irrespective of other multipliers, a cosine operation can only return values between  and . To think of this a different way,  will give us the same returns as , only we will move around the unit circle five times as much before finding our answer.

Thus, our final range is .

The range of the function represents the spread of possible answers you can get for , given all values of . In this case, the ordinary range for a cosine function is , since the largest value that cosine can solve to is  (for a cosine of  or a multiple of one of those values), and the smallest value cosine can solve to is  (for a cosine of  or a multiple of one of those values).

One fast way to match a range to a function is to look for the function which has a vertical shift equal to the mean of the range values. In other words, for the standard trigonometric function , where  represents the vertical shift, .

In this case, since our range is , we expect our  to be .

Of the answer choices, only  has , so we know this is our correct choice.

The range of the function represents the spread of possible answers you can get for , given all values of . In this case, the ordinary range for a cosine function is , since the largest value that cosine can solve to is  (for a cosine of  or a multiple of one of those values), and the smallest value cosine can solve to is  (for a cosine of  or a multiple of one of those values).

One fast way to match a range to a function is to look for the function which has a vertical shift equal to the mean of the range values. In other words, for the standard trigonometric function , where  represents the vertical shift, .

In this case, since our range is , we expect our  to be .

Of the answer choices, only  has , so we know this is our correct choice.

The range of the function represents the spread of possible answers you can get for , given all values of . In this case, the ordinary range for a cosine function is , since the largest value that cosine can solve to is  (for a cosine of  or a multiple of one of those values), and the smallest value cosine can solve to is  (for a cosine of  or a multiple of one of those values).

One fast way to match a range to a function is to look for the function which has a vertical shift equal to the mean of the range values. In other words, for the standard trigonometric function , where  represents the vertical shift, .

In this case, since our range is , we expect our  to be .

Of the answer choices, only  has , so we know this is our correct choice.

The range of a cosine 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 , or , 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.

Based on this data, we can make a little triangle that looks like:

This is because .

Now, this means that  must equal .  (Recall that the cosine function is negative in the second quadrant.) Now, we are looking for:

 or .  This is the cosine of a reference angle of:

Looking at our little triangle above, we can see that the cosine of  is .

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.")  

Now, it is easiest to think of this like you are drawing a little triangle in the third quadrant of the Cartesian plane. It would look like:

So, you first need to  calculate the hypotenuse. You can do this by using the Pythagorean Theorem, , where  and  are the lengths of the legs of the triangle and  the length of the hypotenuse. Rearranging the equation to solve for , you get:

Substituting in the given values:

So, the cosine of an angle is:

  or, for your data, .  

This is approximately . Rounding, this is . However, since  is in the third quadrant your value must be negative: .

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.") Now, it is easiest to think of this like you are drawing a little triangle in the second quadrant of the Cartesian plane. It would look like:

So, you first need to  calculate the hypotenuse:

So, the cosine of an angle is:

  or, for your data, .  

This is approximately . Rounding, this is . However, since  is in the second quadrant your value must be negative: .

If the point to be reached is , then we may envision a right triangle with sides  and , and hypotenuse . The Pythagorean Theorem tells us that , so we plug in and find that: 

Thus, 

Now, SOHCAHTOA tells us that , so we know that:

Thus, our cosine is approximately . However, as we are in the third quadrant, cosine must be negative! Therefore, our true cosine is .

If the point to be reached is , then we may envision a right triangle with sides  and , and hypotenuse . The Pythagorean Theorem tells us that , so we plug in and find that: .

Thus, .

Now, SOHCAHTOA tells us that , so we know that:

Thus, our cosine is approximately . However, as we are in the second quadrant, cosine must be negative! Therefore, our true cosine is .