There are known values for a specific set of trigonometric functions within all quadrants that follows a pattern.
Trigonometric Functions
sinθ=oppositehypotenuse
cosθ=adjacenthypotenuse
tanθ=oppositeadjacent
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
1st Quadrant(Q1)
=
(+,+)
2nd Quadrant(Q2)
=
(-,+)
3rd Quadrant(Q3)
=
(-,-)
4th Quadrant(Q4)
=
(+,-)
Given that (cosθ,sinθ)=(x,y), sin60° will be the y-coordinate of 60°, which is √32
sin60°=√32
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°
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°=π
Given that value, we can easily compute for the radian values of 30°,45°, and 60°
30°×π180°
=
30°π180°
=
π6
45°×π180°
=
45°π180°
=
π4
60°×π180°
=
60°π180°
=
π3
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)π3
=
2π3
135°∣∣45°
135°
=
(4-1)π4
=
3π4
150°∣∣30°
150°
=
(6-1)π6
=
5π6
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)π6
=
7π6
225°↔45°
225°
=
(1+4)π4
=
5π4
240°↔60°
240°
=
(1+3)π3
=
4π3
300°↔120°
300°
=
(2+3)π3
=
5π3
315°↔135°
315°
=
(3+4)π4
=
7π4
330°↔150°
330°
=
(5+6)π6
=
11π6
Given these known values, the radian value of 330° would be 11π6
330°=11π6
Method Two
Since we know that 180°=π, we can easily get the radian value of 1°
180°
=
π
180°180°
=
π180°
Divide both sides by 180°
1°
=
π180°
Now that we know the radian value of 1°, we can multiply 330° to it in order to get its radian value.
Using the given image, we can see that the radius of the circle will be 1. Therefore, the circle follows the formula x2+y2=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 θ, 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
sin
=
oppositehypotenuse
=
y1
=
y
cos
=
adjacenthypotenuse
=
x1
=
x
tan
=
oppositeadjacent
=
yx
x and y are both coordinate values
Given the following values, we can use the image as reference and solve for sin90°,cos180°, and tan360°
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
degrees
=
radians×180°π
=
7π6×180°π
=
1260°6
ππ=1
=
210°
Next, recall the values of the trigonometric functions
Given that (cosθ,sinθ)=(x,y), we are given the following values
sin210°
=
-12
cos210°
=
-√32
Finally, solve for the value of tan210°
tan210°
=
sin210°cos210°
=
-12-√32
Substitute known values
=
1√3
Simplify
tan(7π6)
=
1√3
tan(7π6)=1√3
Method Two
First, convert the radian to degrees
degrees
=
radians×180°π
=
7π6×180°π
=
1260°6
ππ=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 3rd quadrant, which means tan210° is positive.
To find θ, we can subtract 180° (the horizontal line) from 210°.
210°-180°=30°
This means we can also refer to tan(7π6) 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°
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
degrees
=
radians×180°π
=
5π3×180°π
=
900°3
ππ=1
=
300°
Alternatively, we can convert the value to a mixed fraction.
5π3
=
123π
Knowing these values, we can identify the quadrant that contains 300° or 5π3 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 π and 23 to 2π
The value should be located in the fourth quadrant.
Now use a special triangle to find the value of 5π3
Note that 5π3 can be written as 5×π3 so we can use the π3 value as a reference angle.
tan300°
=
oppositeadjacent
=
√31
Substitute known values
=
√3
Simplify
Finally, keep the following in mind when providing the proper value of trigonometric functions for each quadrant
1st Quadrant
=
All are positive
2nd Quadrant
=
Only sin is positive
3rd Quadrant
=
Only tan is positive
4th Quadrant
=
Only cos is positive
Therefore, the known value of tan5π3=-√3
tan(5π3)=-√3
Method Two
First, convert the radian to degrees
degrees
=
radians×180°π
=
5π3×180°π
=
900°3
ππ=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 4th quadrant, which means tan300° is negative.
To find θ, we can subtract 300° from 360° (the horizontal line).
360°-300°=60°
This means we can also refer to tan(5π3) 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°
=
oppositeadjacent
tan60°
=
√31
tan(5π3)
=
√3
Recall that the angle lies on the 4th quadrant, so the answer should be negative.
There are known values for a specific set of trigonometric functions within all quadrants that follows a pattern.
Trigonometric Functions
sinθ=oppositehypotenuse
cosθ=adjacenthypotenuse
tanθ=oppositeadjacent
Values of a Right Triangle in a Unit Circle
Method One
First, find the acute reference angle by finding the value to be added to π to get 5π4.
π can be written as 4π4. So if we subtract this value from 5π4, we can get the acute reference angle.
5π4-4π4
=
π4
Therefore, 5π4 can be written as π+π4 with π4 as the reference angle.
Next, we can identify the quadrant that contains 5π4 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 5π4
sinπ4
=
oppositehypotenuse
=
1√2
Substitute known values
Finally, keep the following in mind when providing the proper value of trigonometric functions for each quadrant
1st Quadrant
=
All are positive
2nd Quadrant
=
Only sin is positive
3rd Quadrant
=
Only tan is positive
4th Quadrant
=
Only cos is positive
Therefore, the known value of sin5π4=-1√2
sin5π4=-1√2
Method Two
First, convert the radian to degrees
degrees
=
radians×180°π
=
5π4×180°π
=
900°4
ππ=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 3rd quadrant, which means sin225° is negative.
To find θ, we can subtract 180° (the horizontal line) from 225°.
225°-180°=45°
This means we can also refer to sin(5π4) 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°
=
oppositehypotenuse
sin45°
=
1√2
sin(5π4)
=
1√2
Recall that the angle lies on the 3rd quadrant, so the answer should be negative.
There are known values for a specific set of trigonometric functions within all quadrants that follows a pattern.
Trigonometric Functions
sinθ=oppositehypotenuse
cosθ=adjacenthypotenuse
tanθ=oppositeadjacent
Values of a Right Triangle in a Unit Circle
Method One
First, you can convert the radian to degrees
degrees
=
radians×180°π
=
2π3×180°π
=
360°3
ππ=1
=
120°
Knowing these values, we can identify the quadrant that contains 120° or 2π3 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 π to get 2π3.
π can be written as 3π3. So if we subtract 2π3 from this value, we can get the acute reference angle.
3π3-2π3
=
π3
Therefore, 2π3 can be written as π-π3 with π3 as the reference angle..
Now use a special triangle that has the reference angle to find the value of 2π3
cosπ3
=
adjacenthypotenuse
=
12
Substitute known values
Finally, keep the following in mind when providing the proper value of trigonometric functions for each quadrant
1st Quadrant
=
All are positive
2nd Quadrant
=
Only sin is positive
3rd Quadrant
=
Only tan is positive
4th Quadrant
=
Only cos is positive
Therefore, the known value of cos2π3=-12
cos2π3=-12
Method Two
First, convert the radian to degrees
degrees
=
radians×180°π
=
2π3×180°π
=
360°3
ππ=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 2nd quadrant, which means cos120° is negative.
To find θ, we can subtract 120° (the horizontal line) from 180°.
180°-120°=60°
This means we can also refer to cos(2π3) 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°
=
adjacenthypotenuse
cos60°
=
12
cos(2π3)
=
12
Recall that the angle lies on the 2nd quadrant, so the answer should be negative.