This is the figure being described in the problem, and as sine is opposite over hypotenuse and cosine adjacent over hypotenuse, the sine ofand the cosine ofwill yield the correct answer.

Rearrange algebraically so that,

.

Over the interval 0 to 360 degrees, sinx = 1 when x equals 90 degrees.

First, think of the angle values for which. (This is the equivalent of taking the arctan.)

.

The angles for which this is true are 45 degrees and 225 degrees.

We set x-15 equal to those two angles and solve for x, giving us 60 and 240.

Rearrange the equation so that,

.

Recall the angles over the interval 0 to 360 degrees for which sec is equal to 2.

These are 60 and 300 degrees.

Set x+7 equal to these angle measures and then find that x equals 53 and 293.

Rearrange the equation so that,

.

Take the square of both sides and then recall the angle measures for which,

.

These measures over the interval

.

Rearrange the equation so that,

.

Take the square of both sides and find the angles for which

.

These two angles areand.

To solve for x, first set up a trigonometric equation using the information provided in the diagram. The two side lengths given are the hypotenuse, x, and the side opposite the given angle, 6. We can set up our equation like this:

The sine ofis, so we can substitue that in:

cross multiplying gives us.

To solve for x, first set up a trigonometric equation using the information provided in the diagram. The two side lengths given are x, the side opposite the angle, and 3, the side adjacent to the angle. This means we'll be using tangent. Set up the equation like this:

We can't just know the tangent by using the unit circle, but we can easily figure it out using sine and cosine. Tangent can be evaluated as sine over cosine.

The sine ofis, and the cosine is. Find the tangent by dividing:

Now we can substitute that value into the original equation we set up:

multiply both sides by 3

Start by setting up the trigonometric ratio using the angles and sides given. We have the side length adjacent to the angle, 1, and the hypotenuse, x, so we will be using cosine:

The cosine ofis, so this makes our equation now:

. x must equal 2.

Examining the equation of this form:

We can find,,andusing the clues given:

• because the range ofindicates an amplitude of.
• becauseis the midpoint betweenand.
• because the period is not changed from the standard.
• . The combined fact thatand(halfway through the period) isindicates a cosine function would work - only the answer must be given as a sine function. Fortunately, though, we can use a shift-related property of.

All told, once we realize these, then we can see that
fits our criteria.