Solve for all x on the interval 

Remember Soh, Cah, Toa?

For this problem it helps to recall that 

Since our tangent is equal to 1 in this problem, we know that our opposite and adjacent sides must be the same (otherwise we wouldn't get "1" when we divided them)

Can you think of any angles in the first quadrant which yield equal x and y values? 

If you guessed  you guessed right! Remember that your  angle in the unit circle will give you a  triangle, which will have equal height and base.

One can setup the relationship

. 

After taking the arccosine,

the arccosine cancels out the cosine leaving just the value of .

One can setup the relationship

.

After taking the arctangent,

the arctangent cancels out the tangent and we are left with the value of .

If the sine of an angle, in this case  is , the angle must be or .

Then we need to solve for theta by dividing by 3:

Which of the following could be a value of ?

To begin, it will be helpful to recall the following property of tangent:

This means that if  our sine and cosine must have equal absolute values, but with opposite signs.

The only place where we will have equal values for sine and cosine will be at the locations halfway between our quadrantal angles (axes). In other words, our answer will align with one of the  angles. 

Additionally, because our sine and cosine must have opposite signs (one negative and one positive), we need to be in either quadrant 2 or quadrant 4. There is only answer from either of those two, so our answer must be .

Since we know the value of the trigonometric function and the triangle is located in Quadrant I, we can draw the triangle and get a sense of it. If the opposite side is 1 and the hypotenuse is 2, we know that we're dealing with a 30-60-90 special triangle. And since the opposite side of the angle is 1, we know that the angle is .

Find 1 possible value of  Given the following:

Recall that

So if , then 

Thinking back to our unit circle, recall that cosine corresponds to the x-value. Therefore, we must be in quadrants II or III. 

So, which angles correspond to an x-value of -0.5? Well, they must be the angles closest to the y-axis, which are our  increment angles. 

This means our angle must be either 

or 

It must be , because 240 is not an option.

Note that there are technically infinte solutions, because we are not given a specific interval. However, we only need to worry about one.

Given the point on the coordinate plane , the origin to this point can be computed by the Pythagorean Theorem.

The hypotenuse of the right triangle formed by the origin and the point is .

The length of the triangle is 1 unit, and the height of the triangle is 5.

Sine of an angle is opposite side divided by the hypotenuse.

Rationalize the denominator.

First, use the Pythagorean Theorem to solve for all the sides of the triangle. You know that the adjacent side is 4 units long, and the opposite side is -9 units long.

Using the Pythagorean Theorem, you should get a hypotenuse of .

Secant is defined as hypotenuse/opposite.

Thus, giving you an answer of .

This is a 30-60-90 triangle. A 30-60-90 triangle will have leg lengths of  and 1 and a hypotenuse of 2. To find the sin value, you need to divide the opposite leg length with the hypotenuse (opposite/hypotenuse). Thus, giving you .