Find the degree measure of an angle for which the value of a trigonometric function is known - Pre-Calculus

One can setup the relationship

.

After taking the arccosine,

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

Solve for all x on the interval

We can begin by recalling which two quadrants have a positive sine. Because sine corresponds to the y-value, we know that sine is positive in quadrants I and II.

Next, recall where we get .

always corresponds to our -increment angles. In this case, the angles we are looking for are and , because those are the two -increment angles in the first two quadrants.

Now, you might be saying, "what about ? That is an increment of 45."

While that is true, , not

So our answer is:

,

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.

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.

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 .

One can setup the relationship

.

After taking the arctangent,

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

One can setup the relationship

.

After taking the arccosine,

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

Solve for all x on the interval

We can begin by recalling which two quadrants have a positive sine. Because sine corresponds to the y-value, we know that sine is positive in quadrants I and II.

Next, recall where we get .

always corresponds to our -increment angles. In this case, the angles we are looking for are and , because those are the two -increment angles in the first two quadrants.

Now, you might be saying, "what about ? That is an increment of 45."

While that is true, , not

So our answer is:

,

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.

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.

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 .

One can setup the relationship

.

After taking the arctangent,

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

One can setup the relationship

.

After taking the arccosine,

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

Solve for all x on the interval

We can begin by recalling which two quadrants have a positive sine. Because sine corresponds to the y-value, we know that sine is positive in quadrants I and II.

Next, recall where we get .

always corresponds to our -increment angles. In this case, the angles we are looking for are and , because those are the two -increment angles in the first two quadrants.

Now, you might be saying, "what about ? That is an increment of 45."

While that is true, , not

So our answer is:

,

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.