Finding Missing Angles Using the Unit Circle

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Question 1 of 3

1. Question
Given that `0≤theta≤2pi`, solve for `theta`

`tan^2theta=3`
- `pi/3,(2pi)/3,(4pi)/3,(5pi)/3`
- `pi/3,(4pi)/3`
- `pi/6,(5pi)/6,(7pi)/6,(11pi)/6`
- `pi/4,(7pi)/6`
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Trigonometric Functions

$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$

$$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$

$$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$

First, simplify the given equation

`tan^2theta` `=` `3`

`sqrt(tan^2theta)` `=` `sqrt3` Get the square root of both sides

`tan theta` `=` `+-sqrt3`

Remember that `tan theta=(sin theta)/(cos theta)`. Knowing that `tan theta` can be either positive or negative, this means that `theta` lies on all `4` quadrants

Next, find the reference angle to work with by using special triangles

`tan theta` `=` `+-sqrt3` `=` $$\frac{\text{opposite}}{\text{adjacent}}$$

`tan theta` `=` `+-sqrt3` `=` `(sqrt3)/1` Using `pi/3` as `theta`

`tan theta` `=` `+-sqrt3` `=` `sqrt3`

Hence, the reference angle and the value of `theta` on the First Quadrant is `pi/3`

Finally, use the reference angle to find the value of `theta` on the other `3` quadrants

Remember that the reference angle is always relative to the `x`-axis

Second Quadrant:

`theta` `=` `pi-pi/3`

`=` `(3pi)/3-pi/3`

`=` `(2pi)/3`

Third Quadrant:

`theta` `=` `pi+pi/3`

`=` `(3pi)/3+pi/3`

`=` `(4pi)/3`

Fourth Quadrant:

`theta` `=` `2pi-pi/3`

`=` `(6pi)/3-pi/3`

`=` `(5pi)/3`

`theta=pi/3,(2pi)/3,(4pi)/3,(5pi)/3`
Question 2 of 3

2. Question
Given that `0≤theta≤2pi`, solve for `theta`

`sin^2theta=1`
- `pi/2,(3pi)/2`
- `pi/2,pi,(3pi)/2,2pi`
- `pi/4,(3pi)/4`
- `pi/4,(3pi)/4,(5pi)/4,(7pi)/4`
Hint

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Correct

Correct!

Incorrect

Trigonometric Functions

$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$

$$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$

$$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$

First, simplify the given equation

`sin^2theta` `=` `1`

`sqrt(sin^2theta)` `=` `sqrt1` Get the square root of both sides

`sin theta` `=` `+-1`

In the unit circle. mark out the coordinates on the horizontal and vertical axis

Keep in mind that the radius of a unit circle is `1`

Now, recall that in a coordinate, `(x,y)=(cos theta, sin theta)`

`sin theta=+-1`

The `y` value in these coordinates match the value of `sin theta`. Therefore, `theta=pi/2,(3pi)/2`

`theta=pi/2,(3pi)/2`
Question 3 of 3

3. Question
Given that `0≤theta≤2pi`, solve for `theta`

`costheta=-1`
- `theta=pi`
- `theta=pi/2`
- `theta=(3pi)/2`
- `theta=2pi`
Hint

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Correct

Exceptional!

Incorrect

Trigonometric Functions

$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$

$$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$

$$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$

In the unit circle. mark out the coordinates on the horizontal and vertical axis

Keep in mind that the radius of a unit circle is `1`

Now, recall that in a coordinate, `(x,y)=(cos theta, sin theta)`

`cos theta=-1`

The `x` value in `pi` the value of `cos theta`. Therefore, `theta=pi`

`theta=pi`

Finding Missing Angles Using the Unit Circle

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Quiz summary

Information

1. Question

Hint

Excellent!

Trigonometric Functions

2. Question

Hint

Correct!

Trigonometric Functions

3. Question

Hint

Exceptional!

Trigonometric Functions

Quizzes

`tan^2theta`	`=`	`3`
`sqrt(tan^2theta)`	`=`	`sqrt3`	Get the square root of both sides
`tan theta`	`=`	`+-sqrt3`

`tan theta`	`=`	`+-sqrt3`	`=`	$$\frac{\text{opposite}}{\text{adjacent}}$$

`tan theta`	`=`	`+-sqrt3`	`=`	`(sqrt3)/1`	Using `pi/3` as `theta`

`tan theta`	`=`	`+-sqrt3`	`=`	`sqrt3`

`sin^2theta`	`=`	`1`
`sqrt(sin^2theta)`	`=`	`sqrt1`	Get the square root of both sides
`sin theta`	`=`	`+-1`