The Unit Circle

Question 1 of 7

Find the value of

sin 60 °

$sin60°$

1. $\frac{1}{2}$ $1/2$
2. $\frac{\sqrt{3}}{2}$ $(sqrt(3))/(2)$
3. $\frac{\sqrt{2}}{3}$ $(sqrt(2))/(3)$
4. $- \frac{1}{2}$ $-1/2$

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Coordinates of Trigonometric Functions

(cos θ, sin θ) = (x, y)

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

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

Trigonometric Functions

sin θ = \frac{opposite}{hypotenuse}

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

cos θ = \frac{adjacent}{hypotenuse}

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

tan θ = \frac{opposite}{adjacent}

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

Method One

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

2

$2$

The x-coordinate’s numerator will be the equivalent of the square root of

1, 2,

$1, 2,$ and

3

$3$ 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,

$1, 2,$ and

3

$3$ 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

$1st Quadrant (Q 1)$ $\text(1st Quadrant) (Q1)$	$=$ $=$	$(+, +)$ $(+,+)$
$2nd Quadrant (Q 2)$ $\text(2nd Quadrant) (Q2)$	$=$ $=$	$(-, +)$ $(-,+)$
$3rd Quadrant (Q 3)$ $\text(3rd Quadrant) (Q3)$	$=$ $=$	$(-, -)$ $(-,-)$
$4th Quadrant (Q 4)$ $\text(4th Quadrant) (Q4)$	$=$ $=$	$(+, -)$ $(+,-)$

Given that

(

$($ $cos θ$ $\text(cos)theta$

,

$,$ $sin θ$ $\text(sin)theta$

) = (

$)=($ $x$ $x$

,

$,$ $y$ $y$

)

$)$ ,

sin 60 °

$\text(sin)60°$ will be the y-coordinate of

60 °

$60°$ , which is $\frac{\sqrt{3}}{2}$ $(sqrt(3))/(2)$

sin 60 ° = \frac{\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 °

$60°$ , we can use the

30 - 60 - 90

$30-60-90$ triangle.

To solve for

sin 60 °

$sin60°$ , 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

sin 60 °

$sin60°$

$sin 60 °$ $sin60°$	$=$ $=$	$\frac{\color{#004ec4}{\text{opposite}}}{\color{#e85e00}{\text{hypotenuse}}}$

$sin 60 °$ $sin60°$	$=$ $=$	$\frac{\color{#004ec4}{\sqrt{3}}}{\color{#e85e00}{2}}$

sin 60 ° = \frac{\sqrt{3}}{2}

$\text(sin)60°=(sqrt(3))/(2)$

Question 2 of 7

Find the radian value of

330 °

$330°$

1. $\frac{7 π}{4}$ $(7pi)/4$
2. $\frac{3 π}{4}$ $(3pi)/4$
3. $\frac{5 π}{4}$ $(5pi)/4$
4. $\frac{11 π}{6}$ $(11pi)/6$

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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

$180°=pi$

Given that value, we can easily compute for the radian values of

30 °, 45 °,

$30°, 45°,$ and

60 °

$60°$

$30 ° \times \frac{π}{180} °$ $30°xx(pi)/180°$	$=$ $=$	$\frac{30 ° π}{180} °$ $(30°pi)/180°$

	$=$ $=$	$\frac{π}{6}$ $(pi)/6$

$45 ° \times \frac{π}{180} °$ $45°xx(pi)/180°$	$=$ $=$	$\frac{45 ° π}{180} °$ $(45°pi)/180°$

	$=$ $=$	$\frac{π}{4}$ $(pi)/4$

$60 ° \times \frac{π}{180} °$ $60°xx(pi)/180°$	$=$ $=$	$\frac{60 ° π}{180} °$ $(60°pi)/180°$

	$=$ $=$	$\frac{π}{3}$ $(pi)/3$

For the second quadrant, you can get the radian value of

120 °, 135 °,

$120°, 135°,$ and

150 °

$150°$ 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 °

$120°||60°$

$120 °$ $120°$	$=$ $=$	$\frac{(3 - 1) π}{3}$ $((3-1)pi)/3$

	$=$ $=$	$\frac{2 π}{3}$ $(2pi)/3$

135 ° ∣ ∣ 45 °

$135°||45°$

$135 °$ $135°$	$=$ $=$	$\frac{(4 - 1) π}{4}$ $((4-1)pi)/4$

	$=$ $=$	$\frac{3 π}{4}$ $(3pi)/4$

150 ° ∣ ∣ 30 °

$150°||30°$

$150 °$ $150°$	$=$ $=$	$\frac{(6 - 1) π}{6}$ $((6-1)pi)/6$

	$=$ $=$	$\frac{5 π}{6}$ $(5pi)/6$

For the third and fourth quadrant, you can get the radian value of

210 °, 225 °, 240 °, 300 °, 315 °

$210°, 225°, 240°, 300°, 315°$ and

330 °

$330°$ 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 ° \leftrightarrow 30 °

$210°↔30°$

$210 °$ $210°$	$=$ $=$	$\frac{(1 + 6) π}{6}$ $((1+6)pi)/6$

	$=$ $=$	$\frac{7 π}{6}$ $(7pi)/6$

225 ° \leftrightarrow 45 °

$225°↔45°$

$225 °$ $225°$	$=$ $=$	$\frac{(1 + 4) π}{4}$ $((1+4)pi)/4$

	$=$ $=$	$\frac{5 π}{4}$ $(5pi)/4$

240 ° \leftrightarrow 60 °

$240°↔60°$

$240 °$ $240°$	$=$ $=$	$\frac{(1 + 3) π}{3}$ $((1+3)pi)/3$

	$=$ $=$	$\frac{4 π}{3}$ $(4pi)/3$

300 ° \leftrightarrow 120 °

$300°↔120°$

$300 °$ $300°$	$=$ $=$	$\frac{(2 + 3) π}{3}$ $((2+3)pi)/3$

	$=$ $=$	$\frac{5 π}{3}$ $(5pi)/3$

315 ° \leftrightarrow 135 °

$315°↔135°$

$315 °$ $315°$	$=$ $=$	$\frac{(3 + 4) π}{4}$ $((3+4)pi)/4$

	$=$ $=$	$\frac{7 π}{4}$ $(7pi)/4$

330 ° \leftrightarrow 150 °

$330°↔150°$

$330 °$ $330°$	$=$ $=$	$\frac{(5 + 6) π}{6}$ $((5+6)pi)/6$

	$=$ $=$	$\frac{11 π}{6}$ $(11pi)/6$

Given these known values, the radian value of

330 °

$330°$ would be

\frac{11 π}{6}

$(11pi)/6$

330 ° = \frac{11 π}{6}

$330°=(11pi)/6$

Method Two

Since we know that

180 ° = π

$180°=pi$ , we can easily get the radian value of

1 °

$1°$

$180 °$ $180°$	$=$ $=$	$π$ $pi$

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

$1 °$ $1°$	$=$ $=$	$\frac{π}{180 °}$ $pi/(180°)$

Now that we know the radian value of

1 °

$1°$ , we can multiply

330 °

$330°$ to it in order to get its radian value.

$1 °$ $1°$	$=$ $=$	$\frac{π}{180} °$ $pi/180°$

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

$330 °$ $330°$	$=$ $=$	$330 ° \frac{π}{180 °}$ $330°pi/(180°)$

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

$330°$	$=$ $=$	$\frac{11\pi}{6}$

330 ° = \frac{11 π}{6}

$330°=(11pi)/6$

Question 3 of 7

Given the image below, find the value of:

(i) sin 90 °

$(i) \text(sin)90°$

(i i) cos 180 °

$(ii) \text(cos)180°$

(i i i) tan 360 °

$(iii) \text(tan)360°$

$(i) sin 90 ° =$ $(i) \text(sin)90°=$ (1)

$(i i) cos 180 ° =$ $(ii) \text(cos)180°=$ (-1)

$(i i i) tan 360 ° =$ $(iii) \text(tan)360°=$ (0)

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Trigonometric Functions

sin θ = \frac{opposite}{hypotenuse}

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

cos θ = \frac{adjacent}{hypotenuse}

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

tan θ = \frac{opposite}{adjacent}

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

Using the given image, we can see that the radius of the circle will be

1

$1$ . Therefore, the circle follows the formula

x^{2} + y^{2} = 1

$x^2+y^2=1$ , where

1

$1$ is the radius.

Make a reference triangle by making a line from the point of origin

(0, 0)

$(0,0)$ 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

$theta$ , 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

$1$ as the hypotenuse, we can use the trigonometric functions to find their respective values

$sin$ $\text(sin)$	$=$ $=$	$\frac{opposite}{hypotenuse}$ $\text(opposite)/(\text(hypotenuse))$

	$=$	$y/1$

	$=$	$y$

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

	$=$	$x/1$

	$=$	$x$

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

	$=$	$y/x$

$x$ 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

Find the value of

$tan((7pi)/6)$

1. $tan((7pi)/6)=-sqrt3$
2. $tan((7pi)/6)=-1/(sqrt3)$
3. $tan((7pi)/6)=sqrt3$
4. $tan((7pi)/6)=1/(sqrt3)$

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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 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

Find the value of

$tan((5pi)/3)$

1. $tan((5pi)/3)=-1/(sqrt3)$
2. $tan((5pi)/3)=sqrt3$
3. $tan((5pi)/3)=-sqrt3$
4. $tan((5pi)/3)=1/(sqrt3)$

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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 One

First, 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 Two

First, 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

Find the value of

$sin((5pi)/4)$

1. $\text(sin)(5pi)/4=-(sqrt2)$
2. $\text(sin)(5pi)/4=-1/(sqrt2)$
3. $\text(sin)(5pi)/4=1/(sqrt2)$
4. $\text(sin)(5pi)/4=(sqrt2)$

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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 One

First, 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

$\text(sin)(5pi)/4=-1/(sqrt2)$

Method Two

First, 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

Find the value of

$cos((2pi)/3)$

1. $\text(cos)(2pi)/3=-1/2$
2. $\text(cos)(2pi)/3=2$
3. $\text(cos)(2pi)/3=1/2$
4. $\text(cos)(2pi)/3=-2$

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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 One

First, 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

$\text(cos)(2pi)/3=-1/2$

Method Two

First, 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.

$cos((2pi)/3)$

$=$

$-\frac{1}{2}$

$\text(cos)(2pi)/3=-1/2$

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Quiz summary

Information

1. Question

Hint

Great Work!

Coordinates of Trigonometric Functions

Trigonometric Functions

2. Question

Hint

Correct!

3. Question

Hint

Keep Going!

Trigonometric Functions

4. Question

Hint

Fantastic!

Converting Radian to Degrees

Coordinates of Trigonometric Functions

Trigonometric Functions

5. Question

Hint

Excellent!

Converting Radian to Degrees

Coordinates of Trigonometric Functions

Trigonometric Functions

6. Question

Hint

Nice Job!

Coordinates of Trigonometric Functions

Trigonometric Functions

Values of a Right Triangle in a Unit Circle

7. Question

Hint

Well Done!

Coordinates of Trigonometric Functions

Trigonometric Functions

Values of a Right Triangle in a Unit Circle

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