Values on the Unit Circle

Years

Try VividMath Premium to unlock full access

Start Free Trial

Time limit: 0

Quiz summary

0 of 7 questions completed

Questions:

Information

–

You have already completed the quiz before. Hence you can not start it again.

Quiz is loading...

You must sign in or sign up to start the quiz.

You have to finish following quiz, to start this quiz:

Lets see your results!

Points

Score

Time

Retry Quiz

Answered
Review

Question 1 of 7

1. Question
Find the value of `sin60°`
- `(sqrt(3))/(2)`
- `(sqrt(2))/(3)`
- `1/2`
- `-1/2`
Hint

Help Video

Correct

Great Work!

Incorrect

Coordinates of Trigonometric Functions

$$(\color{#00880A}{\text{cos}\theta},\color{#9a00c7}{\text{sin}\theta})=(\color{#00880A}{x},\color{#9a00c7}{y})$$

There are known values for a specific set of trigonometric functions within all quadrants that follows a pattern.

Trigonometric Functions

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

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

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

Method One

The coordinates of each specific value will be a fraction with a denominator of `2`

The x-coordinate’s numerator will be the equivalent of the square root of `1, 2,` and `3` respectively for each quadrant, starting from the value nearest to the y-axis to the value nearest to the x-axis

The y-coordinate’s numerator will be the equivalent of the square root of `1, 2,` and `3` respectively for each quadrant, starting from the value nearest to the x-axis to the value nearest to the y-axis

Remember to apply the proper signs for each value depending on the quadrant they are in

`\text(1st Quadrant) (Q1)` `=` `(+,+)`

`\text(2nd Quadrant) (Q2)` `=` `(-,+)`

`\text(3rd Quadrant) (Q3)` `=` `(-,-)`

`\text(4th Quadrant) (Q4)` `=` `(+,-)`

Given that `(``\text(cos)theta``,``\text(sin)theta``)=(``x``,``y``)`, `\text(sin)60°` will be the y-coordinate of `60°`, which is `(sqrt(3))/(2)`

`\text(sin)60°=(sqrt(3))/(2)`

Method Two

We can use special triangles to solve this problem.

Since the given angle measures `60°`, we can use the `30-60-90` triangle.

To solve for `sin60°`, we can use the known values of the side opposite to it and the hypotenuse.

Since we have the opposite and hypotenuse values, we can solve for `sin60°`

`sin60°` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$

`sin60°` `=` $$\frac{\color{#004ec4}{\sqrt{3}}}{\color{#e85e00}{2}}$$

`\text(sin)60°=(sqrt(3))/(2)`

Question 2 of 7

2. Question

Find the radian value of `330°`

`(11pi)/6`
`(5pi)/4`
`(3pi)/4`
`(7pi)/4`

Hint

Help Video

Correct

Correct!

Incorrect

There are known values for a specific set of trigonometric functions within all quadrants that follows a pattern.

Method One

Keep in mind that `180°=pi`

Given that value, we can easily compute for the radian values of `30°, 45°,` and `60°`

`30°xx(pi)/180°`	`=`	`(30°pi)/180°`

	`=`	`(pi)/6`

`45°xx(pi)/180°`	`=`	`(45°pi)/180°`

	`=`	`(pi)/4`

`60°xx(pi)/180°`	`=`	`(60°pi)/180°`

	`=`	`(pi)/3`

For the second quadrant, you can get the radian value of `120°, 135°,` and `150°` by using the same radian value of their parallels on the first quadrant and changing the numerator’s constant to the difference of their numerator and denominator

`120°||60°`

`120°`	`=`	`((3-1)pi)/3`

	`=`	`(2pi)/3`

`135°||45°`

`135°`	`=`	`((4-1)pi)/4`

	`=`	`(3pi)/4`

`150°||30°`

`150°`	`=`	`((6-1)pi)/6`

	`=`	`(5pi)/6`

For the third and fourth quadrant, you can get the radian value of `210°, 225°, 240°, 300°, 315°` and `330°` by using the same radian value of their opposites on the first and second quadrant and adding the value of their denominator to their numerator’s constant

`210°↔30°`

`210°`	`=`	`((1+6)pi)/6`

	`=`	`(7pi)/6`

`225°↔45°`

`225°`	`=`	`((1+4)pi)/4`

	`=`	`(5pi)/4`

`240°↔60°`

`240°`	`=`	`((1+3)pi)/3`

	`=`	`(4pi)/3`

`300°↔120°`

`300°`	`=`	`((2+3)pi)/3`

	`=`	`(5pi)/3`

`315°↔135°`

`315°`	`=`	`((3+4)pi)/4`

	`=`	`(7pi)/4`

`330°↔150°`

`330°`	`=`	`((5+6)pi)/6`

	`=`	`(11pi)/6`

Given these known values, the radian value of `330°` would be `(11pi)/6`

`330°=(11pi)/6`

Method Two

Since we know that `180°=pi`, we can easily get the radian value of `1°`

`180°`	`=`	`pi`

$$\frac{180°}{\color{#CC0000}{180°}}$$	`=`	$$\frac{\pi}{\color{#CC0000}{180°}}$$	Divide both sides by `180°`

`1°`	`=`	`pi/(180°)`

Now that we know the radian value of `1°`, we can multiply `330°` to it in order to get its radian value.

`1°`	`=`	`pi/180°`

$$1°\times\color{#CC0000}{330°}$$	`=`	$$\frac{\pi}{180°}\times\color{#CC0000}{330°}$$	Multiply both sides by `330°`

`330°`	`=`	`330°pi/(180°)`

$$330°$$	`=`	$$\frac{330°\pi\div\color{#CC0000}{30°}}{180\div\color{#CC0000}{30°}}$$	Divide the numerator and denominator by `30°`

$$330°$$	`=`	$$\frac{11\pi}{6}$$

`330°=(11pi)/6`

Question 3 of 7

3. Question
Given the image below, find the value of:

`(i) \text(sin)90°`

`(ii) \text(cos)180°`

`(iii) \text(tan)360°`
- `(i) \text(sin)90°=` (1)
  
  `(ii) \text(cos)180°=` (-1)
  
  `(iii) \text(tan)360°=` (0)
Hint

Help Video

Correct

Keep Going!

Incorrect

Trigonometric Functions

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

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

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

Using the given image, we can see that the radius of the circle will be `1`. Therefore, the circle follows the formula `x^2+y^2=1`, where `1` is the radius.

Make a reference triangle by making a line from the point of origin `(0,0)` going to any point in the circle to represent the radius and connect it to either the x or y-axis

Notice that, given angle `theta`, the circle has the vertical line parallel to the y-axis as its opposite side and the line on the x-axis as the adjacent side

Now that we have the opposite and adjacent sides and the radius `1` as the hypotenuse, we can use the trigonometric functions to find their respective values

`\text(sin)` `=` `\text(opposite)/(\text(hypotenuse))`

`=` `y/1`

`=` `y`

`\text(cos)` `=` `\text(adjacent)/(\text(hypotenuse))`

`=` `x/1`

`=` `x`

`\text(tan)` `=` `\text(opposite)/(\text(adjacent))`

`=` `y/x`

`x` and `y` are both coordinate values

Given the following values, we can use the image as reference and solve for `\text(sin)90°, \text(cos)180°,` and `\text(tan)360°`

`90°=(0,1)`

`\text(sin)90°` `=` `y` `=` `1`

`180°=(-1,0)`

`\text(cos)180°` `=` `x` `=` `-1`

`360°=(1,0)`

`\text(tan)360°` `=` `y/x` `=` `0/1` `=` `0`

`(i) \text(sin)90°=1`

`(ii) \text(cos)180°=-1`

`(iii) \text(tan)360°=0`

Question 4 of 7

4. Question

Find the value of `tan((7pi)/6)`

`tan((7pi)/6)=1/(sqrt3)`
`tan((7pi)/6)=-1/(sqrt3)`
`tan((7pi)/6)=sqrt3`
`tan((7pi)/6)=-sqrt3`

Hint

Help Video

Correct

Fantastic!

Incorrect

Converting Radian to Degrees

`\text(degrees)=\text(radian)xx(180°)/pi`

Coordinates of Trigonometric Functions

$$(\color{#00880A}{\text{cos}\theta},\color{#9a00c7}{\text{sin}\theta})=(\color{#00880A}{x},\color{#9a00c7}{y})$$

Trigonometric Functions

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

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

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

There are known values for a specific set of trigonometric functions within all quadrants that follows a pattern.

Method One

First, convert the radian to degrees

`\text(degrees)`	`=`	`\text(radians)xx(180°)/pi`

	`=`	`(7pi)/6xx(180°)/pi`

	`=`	`(1260°)/6`	`pi/pi=1`

	`=`	`210°`

Next, recall the values of the trigonometric functions

Given that `(``\text(cos)theta``,``\text(sin)theta``)=(``x``,``y``)`, we are given the following values

`\text(sin)210°`	`=`	`-1/2`

`\text(cos)210°`	`=`	`-(sqrt3)/2`

Finally, solve for the value of `\text(tan)210°`

`\text(tan)210°`	`=`	`(\text(sin)210°)/(\text(cos)210°)`

	`=`	`(-1/2)/(-(sqrt3)/2)`	Substitute known values

	`=`	`1/(sqrt3)`	Simplify

`tan((7pi)/6)`	`=`	`1/(sqrt3)`

`tan((7pi)/6)=1/(sqrt3)`

Method Two

First, convert the radian to degrees

`\text(degrees)`	`=`	`\text(radians)xx(180°)/pi`

	`=`	`(7pi)/6xx(180°)/pi`

	`=`	`(1260°)/6`	`pi/pi=1`

	`=`	`210°`

Make a reference triangle by making a line from the point of origin `(0,0)` going to any point in the circle that would represent the angle `210°` and connect it to either the x or y-axis

From here, we can see that `210°` lies on the `3`rd quadrant, which means `tan210°` is positive.

To find `theta`, we can subtract `180°` (the horizontal line) from `210°`.

`210°-180°=30°`

This means we can also refer to `tan((7pi)/6)` as `tan30°`

Knowing that the reference triangle has `30°` and `90°` angles, we can use the special `30-60-90` triangle.

To solve for `tan30°`, we can use the known values of the sides opposite and adjacent to it.

Since we have the opposite and adjacent values, we can solve for `tan30°`

`tan30°`	`=`	$$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{adjacent}}}$$

`tan30°`	`=`	$$\frac{\color{#004ec4}{1}}{\color{#e85e00}{\sqrt{3}}}$$

`tan((7pi)/6)`	`=`	$$\frac{1}{\sqrt{3}}$$

`\text(tan)((7pi)/6)=1/(sqrt3)`

Question 5 of 7

5. Question

Find the value of `tan((5pi)/3)`

`tan((5pi)/3)=-sqrt3`
`tan((5pi)/3)=sqrt3`
`tan((5pi)/3)=-1/(sqrt3)`
`tan((5pi)/3)=1/(sqrt3)`

Hint

Help Video

Correct

Excellent!

Incorrect