The Unit Circle
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Question 1 of 7
1. Question
Find the value of `sin60°`Hint
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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 OneThe 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 TwoWe 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°`Hint
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There are known values for a specific set of trigonometric functions within all quadrants that follows a pattern.Method OneKeep 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 TwoSince 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°`
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`(i) \text(sin)90°=` (1)`(ii) \text(cos)180°=` (-1)`(iii) \text(tan)360°=` (0)
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Trigonometric Functions
$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$$$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$$$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$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 sideNow 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 valuesGiven 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` -
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Question 4 of 7
4. Question
Find the value of `tan((7pi)/6)`Hint
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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 OneFirst, 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 TwoFirst, 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)`Hint
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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 OneFirst, convert the radian to degrees`\text(degrees)` `=` `\text(radians)xx(180°)/pi` `=` `(5pi)/3xx(180°)/pi` `=` `(900°)/3` `pi/pi=1` `=` `300°` Alternatively, we can convert the value to a mixed fraction.`(5pi)/3` `=` `1 2/3 pi` Knowing these values, we can identify the quadrant that contains `300°` or `(5pi)/3` using this chart:Note that if you prefer using the mixed number form of the radian, you will also try getting the estimated location by finding the quadrant beyond `pi` and `2/3` to `2pi`The value should be located in the fourth quadrant.
Now use a special triangle to find the value of `(5pi)/3`Note that `(5pi)/3` can be written as `5xx(pi)/3` so we can use the `pi/3` value as a reference angle.
`\text(tan)300°` `=` `(\text(opposite))/(\text(adjacent))` `=` `(sqrt3)/1` Substitute known values `=` `sqrt3` Simplify Finally, keep the following in mind when providing the proper value of trigonometric functions for each quadrant`\text(1st Quadrant)` `=` `\text(All are positive)` `\text(2nd Quadrant)` `=` `\text(Only sin is positive)` `\text(3rd Quadrant)` `=` `\text(Only tan is positive)` `\text(4th Quadrant)` `=` `\text(Only cos is positive)` Therefore, the known value of `\text(tan)(5pi)/3=-sqrt3``tan((5pi)/3)=-sqrt3`Method TwoFirst, convert the radian to degrees`\text(degrees)` `=` `\text(radians)xx(180°)/pi` `=` `(5pi)/3xx(180°)/pi` `=` `(900°)/3` `pi/pi=1` `=` `300°` 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 `300°` and connect it to either the x or y-axis
From here, we can see that `300°` lies on the `4`th quadrant, which means `tan300°` is negative.To find `theta`, we can subtract `300°` from `360°` (the horizontal line).`360°-300°=60°`
This means we can also refer to `tan((5pi)/3)` as `tan60°`Knowing that the reference triangle has `60°` and `90°` angles, we can use the special `30-60-90` triangle.To solve for `tan60°`, 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 `tan60°``tan60°` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{adjacent}}}$$ `tan60°` `=` $$\frac{\color{#004ec4}{\sqrt{3}}}{\color{#e85e00}{1}}$$ `tan((5pi)/3)` `=` `sqrt3` Recall that the angle lies on the `4`th quadrant, so the answer should be negative.`tan((5pi)/3)` `=` `-sqrt3` `tan((5pi)/3)=-sqrt3` -
Question 6 of 7
6. Question
Find the value of `sin((5pi)/4)`Hint
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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
$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$$$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$$$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$Values of a Right Triangle in a Unit Circle
Method OneFirst, find the acute reference angle by finding the value to be added to `pi` to get `(5pi)/4`.`pi` can be written as `(4pi)/4`. So if we subtract this value from `(5pi)/4`, we can get the acute reference angle.`(5pi)/4-(4pi)/4` `=` `(pi)/4` Therefore, `5pi/4` can be written as `pi+pi/4` with `pi/4` as the reference angle.Next, we can identify the quadrant that contains `(5pi)/4` using this chart:
The value should be located in the third quadrant.
Now use a special triangle that has the reference angle to find the value of `(5pi)/4`
`\text(sin)pi/4` `=` `(\text(opposite))/(\text(hypotenuse))` `=` `1/(sqrt2)` Substitute known values Finally, keep the following in mind when providing the proper value of trigonometric functions for each quadrant`\text(1st Quadrant)` `=` `\text(All are positive)` `\text(2nd Quadrant)` `=` `\text(Only sin is positive)` `\text(3rd Quadrant)` `=` `\text(Only tan is positive)` `\text(4th Quadrant)` `=` `\text(Only cos is positive)` Therefore, the known value of `\text(sin)(5pi)/4=-1/(sqrt2)``\text(sin)(5pi)/4=-1/(sqrt2)`Method TwoFirst, convert the radian to degrees`\text(degrees)` `=` `\text(radians)xx(180°)/pi` `=` `(5pi)/4xx(180°)/pi` `=` `(900°)/4` `pi/pi=1` `=` `225°` 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 `225°` and connect it to either the x or y-axis
From here, we can see that `225°` lies on the `3`rd quadrant, which means `sin225°` is negative.To find `theta`, we can subtract `180°` (the horizontal line) from `225°`.`225°-180°=45°`
This means we can also refer to `sin((5pi)/4)` as `sin45°`Knowing that the reference triangle has `45°` and `90°` angles, we can use the special `45-45-90` triangle.
To solve for `sin45°`, 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 `sin45°``sin45°` `=` $$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$$ `sin45°` `=` $$\frac{\color{#004ec4}{1}}{\color{#e85e00}{\sqrt{2}}}$$ `sin((5pi)/4)` `=` $$\frac{1}{\sqrt{2}}$$ Recall that the angle lies on the `3`rd quadrant, so the answer should be negative.`sin((5pi)/4)` `=` $$-\frac{1}{\sqrt{2}}$$ `\text(sin)(5pi)/4=-1/(sqrt2)` -
Question 7 of 7
7. Question
Find the value of `cos((2pi)/3)`Hint
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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
$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$$$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$$$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$Values of a Right Triangle in a Unit Circle
Method OneFirst, you can convert the radian to degrees`\text(degrees)` `=` `\text(radians)xx(180°)/pi` `=` `(2pi)/3xx(180°)/pi` `=` `(360°)/3` `pi/pi=1` `=` `120°` Knowing these values, we can identify the quadrant that contains `120°` or `(2pi)/3` using this chart:
The value should be located in the second quadrant.
Next, find the acute reference angle by finding the value to be subtracted from `pi` to get `(2pi)/3`.`pi` can be written as `(3pi)/3`. So if we subtract `(2pi)/3` from this value, we can get the acute reference angle.`(3pi)/3-(2pi)/3` `=` `(pi)/3` Therefore, `2pi/3` can be written as `pi-pi/3` with `pi/3` as the reference angle..Now use a special triangle that has the reference angle to find the value of `(2pi)/3``\text(cos)pi/3` `=` `(\text(adjacent))/(\text(hypotenuse))` `=` `1/2` Substitute known values Finally, keep the following in mind when providing the proper value of trigonometric functions for each quadrant`\text(1st Quadrant)` `=` `\text(All are positive)` `\text(2nd Quadrant)` `=` `\text(Only sin is positive)` `\text(3rd Quadrant)` `=` `\text(Only tan is positive)` `\text(4th Quadrant)` `=` `\text(Only cos is positive)` Therefore, the known value of `\text(cos)(2pi)/3=-1/2``\text(cos)(2pi)/3=-1/2`Method TwoFirst, convert the radian to degrees`\text(degrees)` `=` `\text(radians)xx(180°)/pi` `=` `(2pi)/3xx(180°)/pi` `=` `(360°)/3` `pi/pi=1` `=` `120°` 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 `120°` and connect it to either the x or y-axis
From here, we can see that `120°` lies on the `2`nd quadrant, which means `cos120°` is negative.To find `theta`, we can subtract `120°` (the horizontal line) from `180°`.`180°-120°=60°`
This means we can also refer to `cos((2pi)/3)` as `cos60°`Knowing that the reference triangle has `60°` and `90°` angles, we can use the special `30-60-90` triangle.
To solve for `cos60°`, we can use the known values of the side adjacent to it and the hypotenuse.
Since we have the adjacent and hypotenuse values, we can solve for `cos60°``cos60°` `=` $$\frac{\color{#004ec4}{\text{adjacent}}}{\color{#e85e00}{\text{hypotenuse}}}$$ `cos60°` `=` $$\frac{\color{#004ec4}{1}}{\color{#e85e00}{2}}$$ `cos((2pi)/3)` `=` $$\frac{1}{2}$$ Recall that the angle lies on the `2`nd quadrant, so the answer should be negative.`cos((2pi)/3)` `=` $$-\frac{1}{2}$$ `\text(cos)(2pi)/3=-1/2`