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 atand 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 beand the maximum value to be.

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

Among the possible answers, the one that works is:

Theparameter 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 isin the general formula:
Thus we see amplitude of our function isand so the range is:

Recall that.

Therefore, we are looking foror.

Now, this has a reference angle of, but it is in the third quadrant. This means that the value will be negative. The value ofis. 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,, whereandare leg lengths andis 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,, whereandare leg lengths andis 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 thetriangle appears as follows in radians:

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

Since, we know thatmust 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 foris. Therefore, ifis, then for, it will be.

在一个right triangle, for sides a and b, with c being the hypotenuse,. Thus if cos(A) is, then c = 14, and the side adjacent to A is 11. Therefore, the side opposite of angle A is the square root of, which isSince sin is, sin(A) is.

As with all trigonometry problems, begin by considering how you could rearrange the question. They often have hidden easy ways out. So begin by noticing:

Now, you can treatlike it is any standard denominator. Therefore:

Combine your fractions and get:

Now, from our trig identities, we know that, so we can say:

Now, for our triangle, theis. Therefore,