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 of is , 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 of is , so this makes our equation now:

. x must equal 2.

Examining the equation of this form:

We can find , ,  and using the clues given:

• because the range of indicates an amplitude of .
• because is the midpoint between  and .
• because the period is not changed from the standard .
• . The combined fact that and (halfway through the period) is indicates 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.

Combining a good deal of our information - an amplitude of , a -intercept of and a minimum of with no phase shift - means we are looking for a cosine function. In other words, we can start right at...

We also know that because there is a period of (since when . In other words, we can conclude that the function we are looking for is...

Judging from those five given points, we can draw the following clues:

• Amplitude of
• Period of

Also, note that none of the answer choices have a phase shift, meaning that you can instantly start looking for a sine function because of the zero-rise-zero-fall-zero period that is marked by those five points. Furthermore, we can also realize that we are looking for...

...because we need a period of 4 - i.e,

when .

One important thing to realize is that the frequency is the reciprocal of the period. So if the function has a frequency of Hertz (or cycles per second), the period has to be or seconds. Because when , we know we are looking for a equation that includes .

Also, because we have an amplitude of but a minimum of , there must be a shift upwards by units. Only one function fulfills those two criteria and the period criteria:

Don't get scared off by the fact we're doing trig functions! Factor as you normally would. Because our middle term is negative (), we know that the signs inside of our parentheses will be negative. 

This means that  can be factored to  or .

The fastest way to solve this equation is to simply try the three answers.  Plugging in  gives

Our first choice is valid.

Plugging in  gives

However, since  is undefined, this cannot be a valid answer.

Finally, plugging in  gives

Therefore, our third answer choice is not correct, meaning only 1 is correct.

Therefore, 

and that only happens once in the given interval, at , or 45 degrees.

We have .

Now since

This last expression can be written as :

.

This shows the required result.