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Pythagorean theorem SOHCAHTOA Trig functions Simplifying radicals Rationalize the denominatorExact value of a trig function
Here you will learn about finding the exact trigonometric values, including where they are located on the coordinate plane, how you can derive them, and how you can use them to problem solve.
Students first learn about trigonometry in Geometry and expand that knowledge as they progress through Algebra 2 and Precalculus.
What are exact values of trigonometric functions?
Exact trig values refer to the exact or precise trigonometric values for specific angle measurements. They are typically expressed as a fraction or with square roots, which can be calculated without approximation and are associated with the special angle measurements of 0^{\circ}, \, 30^{\circ}, \, 45^{\circ}, \, 60^{\circ} \text {, and } 90^{\circ}.
These angle measurements can sometimes be presented in radians instead of degrees.
The exact values of the trigonometric functions, sine, cosine, tangent, cosecant, secant, and cotangent are derived from the unit circle. The unit circle is sketched on the coordinate plane below and has its center at the origin (0, \, 0) and a radius of 1 unit.
The equation of the unit circle is x^2+y^2=1. The unit circle along with your knowledge of special right triangles provides the exact value of trig ratios at the specific degree measurements.
![[FREE] Trigonometry Worksheet (Grade 9 to 12)](https://thirdspacelearning.com/wp-content/uploads/2023/07/Trigonometry-check-for-understanding-quiz-listing-image-.png)
[FREE] Trigonometry Worksheet (Grade 9 to 12)
Use this quiz to check your grade 9 to 12 students’ understanding of Trigonometry. 15+ questions with answers covering a range of 9th to 12th grade trigonometry topics to identify areas of strength and support!
DOWNLOAD FREE [FREE] Trigonometry Worksheet (Grade 9 to 12)
Use this quiz to check your grade 9 to 12 students’ understanding of Trigonometry. 15+ questions with answers covering a range of 9th to 12th grade trigonometry topics to identify areas of strength and support!
DOWNLOAD FREELet’s first look at a 30-60-90 triangle on the coordinate plane with the unit circle.
Since the unit circle has a radius length of 1, the hypotenuse of the 30-60-90 triangle has a measure of 1 unit. To find the length of x (the long leg of the 30-60-90 triangle), you can use a proportion.
The ratio of the sides of a 30-60-90 triangle are 1\text{:} \, \sqrt{3}\text{:} \, 2 (short leg: long leg: hypotenuse).
To find x, the proportion is:
\cfrac{\text { long leg }}{\text { hypotenuse }}=\cfrac{\sqrt{3}}{2}=\cfrac{x}{1}This means that
\begin{aligned}&\sqrt{3}=2x \\\\ &\cfrac{\sqrt{3}}{2}=x \end{aligned}To find y, the proportion is:
\begin{aligned}&\cfrac{\text{ short leg }}{\text{ hypotenuse }}=\cfrac{1}{2}=\cfrac{y}{1} \\\\ &1=2 y \\\\ &\cfrac{1}{2}=y \end{aligned}Now that you have the value of x and y, let’s take another look at the triangle.
Using SOHCAHTOA, let’s write the trig ratios with respect to the angle that is 30^{\circ} which is the exact value of the six trig ratios of 30^{\circ}.
\begin{aligned}&\sin{30}=\cfrac{\cfrac{1}{2}}{1}=\cfrac{1}{2} \\\\ &\cos{30}=\cfrac{\cfrac{\sqrt{3}}{2}}{1}=\cfrac{\sqrt{3}}{2} \\\\ &\tan{30}=\cfrac{\cfrac{1}{2}}{\cfrac{\sqrt{3}}{2}}=\cfrac{1}{\sqrt{3}}=\cfrac{\sqrt{3}}{3} \end{aligned}Remember from the trigonometric identities that:
\csc\theta=\cfrac{1}{\sin\theta},~\sec\theta=\cfrac{1}{\cos\theta},~\cot\theta=\cfrac{1}{\tan\theta} \begin{aligned}&\csc{30}=1\div\cfrac{1}{2}=1\times\cfrac{2}{1}=2 \\\\ &\sec{30}=1\div\cfrac{\sqrt{3}}{2}=1\times\cfrac{2}{\sqrt{3}}=\cfrac{2}{\sqrt{3}}=\cfrac{2\sqrt{3}}{3} \\\\ &\cot{30}=1\div\cfrac{\sqrt{3}}{3}=1\times\cfrac{3}{\sqrt{3}}=\cfrac{3\sqrt{3}}{3}=\sqrt{3} \end{aligned}Notice how on the graph, the 30-60-90 triangle touches the edge of the unit circle at the point \left(\cfrac{\sqrt{3}}{2}, \, \cfrac{1}{2}\right) which are the x value and y value of the sides of the triangle.
Also, notice how the sine is the y value and the cosine is the x value of the point on the edge of the unit circle.
Now, let’s follow the same process to find the exact values of the six trigonometric functions of 45^{\circ}.
Using the ratio of the sides of a 45-45-90 triangle and the radius of the unit circle being 1, you can use proportions to solve for x and y.
The ratio of the sides of a 45-45-90 triangle are 1\text{:} \, 1\text{:} \, \sqrt{2} (leg:leg:hypotenuse)
\cfrac{\text{leg}}{\text{hypotenuse}}=\cfrac{1}{\sqrt{2}}=\cfrac{x}{1}This means that
\begin{aligned}&1=\sqrt{2}x \\\\ &\cfrac{1}{\sqrt{2}}=x \\\\ &\cfrac{1}{\sqrt{2}}=\cfrac{\sqrt{2}}{2} \end{aligned}Since it is an isosceles triangle, both x and y are the same value which in simplified form is \cfrac{\sqrt{2}}{2}.
Using SOHCAHTOA, let’s write the trig ratios with respect to the angle that is 45^{\circ} which is the exact value of the six trig ratios of 45^{\circ}.
\begin{aligned}&\sin{45}=\cfrac{\sqrt{2}}{2}\div{1}=\cfrac{\sqrt{2}}{2} \\\\ &\cos{45}=\cfrac{\sqrt{2}}{2}\div{1}=\cfrac{\sqrt{2}}{2} \\\\ &\tan{45}=\cfrac{\sqrt{2}}{2}\div\cfrac{\sqrt{2}}{2}=1 \end{aligned}Remember from the trigonometric identities that:
\csc\theta=\cfrac{1}{\sin\theta},~\sec\theta=\cfrac{1}{\cos\theta},~\cot\theta=\cfrac{1}{\tan\theta} \begin{aligned}&\csc{45}=1\div\cfrac{\sqrt{2}}{2}=1\times\cfrac{2}{\sqrt{2}}=\cfrac{2}{\sqrt{2}}=\sqrt{2} \\\\ &\sec{45}=1\div\cfrac{\sqrt{2}}{2}=1\times\cfrac{2}{\sqrt{2}}=\cfrac{2}{\sqrt{2}}=\sqrt{2} \\\\ &\cot{45}=1\div{1}=1 \end{aligned}Notice how on the graph, the 45-45-90 triangle touches the edge of the unit circle at the point \left(\cfrac{\sqrt{2}}{2}, \, \cfrac{\sqrt{2}}{2}\right) which are the x value and y value of the sides of the triangle.
Also, notice how the sine is the y value and the cosine is the x value of the point on the edge of the unit circle.
Once again, let’s follow the same process to find the exact values of the six trigonometric ratios of 60^{\circ}.
Using the ratio of the sides of a 30-60-90 triangle and the radius of the unit circle being 1, you can use proportions to solve for x and y.
The ratio of the sides of a 30-60-90 triangle are 1\text{:} \, \sqrt{3}\text{:} \, 2 (short leg: long leg: hypotenuse).
To find x, the proportion is:
\begin{aligned}&\cfrac{\text { short leg }}{\text { hypotenuse }}=\cfrac{1}{2}=\cfrac{x}{1} \\\\ &\begin{aligned}& 1=2 x \\\\ & \cfrac{1}{2}=x \end{aligned} \end{aligned}To find y, the proportion is:
\begin{aligned}& \cfrac{\text { long leg }}{\text { hypotenuse }}=\cfrac{\sqrt{3}}{2}=\cfrac{y}{1} \\\\ & \sqrt{3}=2 y \\\\ & \cfrac{\sqrt{3}}{2}=y \end{aligned}
Using SOHCAHTOA, let’s write the trig ratios with respect to the angle that is 60^{\circ} which is the exact value of the six trig ratios of 60^{\circ}.
\begin{aligned}\sin{60}&=\cfrac{\sqrt{3}}{2}\div{1}=\cfrac{\sqrt{3}}{2} \\\\ \cos{60}&=\cfrac{1}{2}\div{1}=\cfrac{1}{2} \\\\ \tan{60}&=\cfrac{\sqrt{3}}{2}\div\cfrac{1}{2}=\cfrac{\sqrt{3}}{2}\times\cfrac{2}{1}=\sqrt{3} \\\\ \csc{60}&=1\div\cfrac{\sqrt{3}}{2}=1\times\cfrac{2}{\sqrt{3}}=\cfrac{2}{\sqrt{3}}=\cfrac{2\sqrt{3}}{3} \\\\ \sec{60}&=1\div\cfrac{1}{2}=1\times\cfrac{2}{1}=2 \\\\ \cot{60}&=1\div\sqrt{3}=\cfrac{1}{\sqrt{3}}=\cfrac{\sqrt{3}}{3} \end{aligned}Notice how on the graph, the 30-60-90 triangle touches the edge of the unit circle at the point \left(\cfrac{1}{2}, \, \cfrac{\sqrt{3}}{2}\right) which are the x value and y value of the sides of the triangle.
Also, notice how the sine is the y value and the cosine is the x value of the point on the edge of the unit circle.
This is true for all the points on the edge of the unit circle.
Let’s find more exact values using the unit circle,
\sin{90} is equal to the y coordinate of the point at 90^{\circ}, so \sin{90}=1 and the \cos{90} is equal to the x coordinate of the point at 90^{\circ}, so in this case, \cos{90}=0. If you know the sine and cosine value, you can easily find the other 4 trig ratios.
\begin{aligned}&\tan{90}=\cfrac{y}{x}=\cfrac{1}{0}=\text { undefined } \\\\ &\csc{90}=1 \\\\ &\sec{90}=\text { undefined } \\\\ &\cot{90}=0 \end{aligned}You can use the same process to find the sine and cosine of 180^{\circ}, \, 270^{\circ}, \, 360^{\circ} and 0^{\circ}.
It’s helpful to organize the exact values of the trig ratios on a table to help you remember them, like the table below. You only have to include the sine, cosine, and tangent on the table since the other trig ratios are the reciprocals of those.
Furthermore, you can use reference angles to find exact values in other quadrants. For example, you can find the exact value of the six trig functions of 135^{\circ} because 135^{\circ} has the same values as 45^{\circ} just in Quadrant II.
This is because 45^{\circ} is the reference angle to 135^{\circ}. The coordinates of 135^{\circ} on the unit circle are \left(- \, \cfrac{\sqrt{2}}{2}, \, \cfrac{\sqrt{2}}{2}\right) which means the \sin{135}=\cfrac{\sqrt{2}}{2} and \cos{135}=- \, \cfrac{\sqrt{2}}{2}.
You can also use the values in the table to find the exact values of trig functions where the angles are not in the first quadrant by using the reference angle.
You can use this table to help figure out which reference angle to use.
Also, the sign value of the angles can be found by the following:
Let’s find the exact value of \cos{240}.
First, since 240^{\circ} is in Quadrant III and not in Quadrant I, let’s find its reference angle. To find the reference angle of an angle in Quadrant III, take the angle and subtract 180^{\circ} from it.
240-180=60^{\circ}The reference angle is 60^{\circ} which means \cos{240} is similar to the value of \cos{60}.
You have already found that \cos{60}=\cfrac{1}{2}. Since 240^{\circ} is in Quadrant III, the cosine value is negative because only the tangent and the cotangent are positive in Quadrant III.
So, the exact value of \cos 240^{\circ}=- \, \cos{60}=- \, \cfrac{1}{2}
Let’s find the exact value of \tan{150}.
150^{\circ} is in Quadrant II, so to find the reference angle, subtract 150^{\circ} from 180^{\circ}.
180-150=30^{\circ}, so 30^{\circ} is the reference angle.
\tan 30^{\circ}=\cfrac{\sqrt{3}}{3}, but since 150^{\circ} is in Quadrant II, the tangent value is negative, so \tan{150}=- \, \tan{30}=- \, \cfrac{\sqrt{3}}{3}
How to find the exact value of a trig function?
How to find the exact value of a trig function?
Common Core State Standards
How does this apply to high school math?
- High School Functions – Trigonometric Functions: (HSF-TF.A.3)
(+) Use special triangles to determine geometrically the values of sine, cosine, tangent for \cfrac{\pi}{3}, \, \cfrac{\pi}{4} and \cfrac{\pi}{6}, and use the unit circle to express the values of sine, cosine, and tangent for \pi-x, \, \pi+x, and 2\pi-x in terms of their values for x, where x is any real number.
How to answer questions involving exact trig values
In order to answer questions involving exact trig values:
- If necessary, find the reference angle.
- Find the value of the trig function using the table or unit circle.
- Determine the sign of the value.
- Write the answer.
Exact trig values examples
Example 1: stating the value
Find the exact value of \cos{30}.
- If necessary, find the reference angle.
30^{\circ} is a reference angle.
2Find the value of the trig function using the table or unit circle.
Using the table or the unit circle, you can determine that \cos30=\cfrac{\sqrt{3}}{2} because at 30^{\circ} on the unit circle, the x -coordinate is \cfrac{\sqrt{3}}{2}.
3Determine the sign of the value.
Since 30^{\circ} is in Quadrant I and a reference angle, the cosine is positive in Quadrant I.
4Write the answer.
\cos{30}=\cfrac{\sqrt{3}}{2}Example 2: stating the value
Find the exact value of \sin{90}.
If necessary, find the reference angle.
90^{\circ} is an angle in the direction of the y -axis, so you do not need to find a reference angle.
Find the value of the trig function using the table or unit circle.
Using the unit circle the \sin{90}=1 because the coordinate at 90^{\circ} is (0, \, 1) and the sine is the same value as the y -coordinate.
Determine the sign of the value.
Since 90^{\circ} is on the y -axis and the point on the unit circle is (0, \, 1), which is positive, the answer is positive.
Write the answer.
Example 3: using the exact values
Find the exact value of \cot{210}.
If necessary, find the reference angle.
The angle is 210^{\circ} which is in Quadrant III, so to find the reference angle subtract 180^{\circ} from 210^{\circ}.
210-180=30^{\circ}
The reference angle is 30^{\circ}.
Find the value of the trig function using the table or unit circle.
The value of \cot{30} is the reciprocal of \tan{30}. Using the table or the unit circle find the value of \tan{30}.
Since the tangent is equal to \cfrac{y}{x}, you can use the y and x coordinates to find the exact value.
\tan{30}=\cfrac{y}{x}=\cfrac{1}{2}\div\cfrac{\sqrt{3}}{2}=\cfrac{1}{\sqrt{3}}.
The reciprocal of \cfrac{1}{\sqrt{3}} is \cfrac{\sqrt{3}}{1}, so \cot{30}=\sqrt{3} .
Determine the sign of the value.
The tangent and the cotangent are both positive in Quadrant III , so the answer is positive.
Write the answer.
Example 4: using the exact values
Find the exact value of \cos{180^{\circ}}.
If necessary, find the reference angle.
180^{\circ} is on the x -axis, so you do not need to find a reference angle.
Find the value of the trig function using the table or unit circle.
Using the unit circle or the table, find the exact value.
\cos 180^{\circ}=- \, 1 because the cosine value is the same as the x -coordinate.
Determine the sign of the value.
The point on the x -axis is (- \, 1, \, 0) so the cosine is negative.
Write the answer.
Example 5: using the exact values
Find the exact value of \sin{45}+\tan{60}.
If necessary, find the reference angle.
45^{\circ} and 60^{\circ} are both in the first quadrant and are reference angles.
Find the value of the trig function using the table or unit circle.
Using the unit circle or table, the \sin{45}=\cfrac{\sqrt{2}}{2} and the \tan{60}=\cfrac{y}{x}=\cfrac{\sqrt{3}}{2}\div\cfrac{1}{2}=\sqrt{3}.
This is because the y -coordinate at 45^{\circ} is \cfrac{\sqrt{2}}{2} and the x and y coordinates at 60^{\circ} are
\left(\cfrac{1}{2}, \, \cfrac{\sqrt{3}}{2}\right)
Determine the sign of the value.
The sine and the tangent are both positive in Quadrant I , so the value is positive.
Write the answer.
Example 6: using the exact values
Find the exact value of \sec{0}+\sin{270}.
If necessary, find the reference angle.
0^{\circ} is on the x -axis and 270^{\circ} is on the y -axis.
Find the value of the trig function using the table or unit circle.
Using the unit circle, find the \cos 0^{\circ} and then take the reciprocal to find the \sec 0^{\circ} and also to find the \sin 270^{\circ}.
The \cos 0^{\circ} is the same as the x -coordinate, which is 1, so the \cos 0^{\circ}=1
The reciprocal of 1 is 1, so \sec 0^{\circ}=1.
The \sin{270}=- \, 1 because the sine is the same as the y -coordinate at 270^{\circ}.
Determine the sign of the value.
Since the angles are on the x -axis and y -axis, look at the signs of coordinates.
Write the answer.
\sin{270}=- \, 1
\sec{0}+\sin{270}=1+(- \, 1)=0
Teaching tips for how to find the exact value of trig functions
- Use discovery based activities so students can discover the exact values through special right triangles and the unit circle.
- Utilize interactive tools online so that students can investigate strategies on how to find the exact values of trig functions.
Easy mistakes to make
- Not leaving the answer as a radical
If you are asked to find the exact value, that means do not get an approximate decimal answer, leave it as a radical.
- Not remembering the exact sine functions
The exact sine values can be difficult to remember, but there is a pattern.
Here it is.
Just be sure to simplify the exact values.
- Not remembering exact cosine functions
The exact cos values can be difficult to remember, but there is a pattern.
Here it is.
Just be sure to simply the exact values.
Notice that the pattern for sine is the reverse pattern for cosine!
Related trigonometry lessons
Practice exact trig values questions
1. What is the exact value of \cos (60)\text{:}
To find the exact value of \cos 60^{\circ}, you can use the table or the unit circle. 60^{\circ} is in the first quadrant and is a reference angle. The cosine is the same as the x -coordinate of the point at 60^{\circ} which is \cfrac{1}{2}.
This means that \cos{60}=\cfrac{1}{2}
You can also use the table to find the value.
2. What is the exact value of \tan (30)?
To find the exact value of the \tan(30) you can use the unit circle. 30^{\circ} is a first quadrant angle and it is a reference angle. Since the tangent is equal to \cfrac{y}{x} or \cfrac{\sin\theta}{\cos\theta}, you can find the coordinate of the point at 30^{\circ} on the unit circle.
\tan(30)=\cfrac{y}{x}=\cfrac{1}{2}\div\cfrac{\sqrt{3}}{2}=\cfrac{1}{\sqrt{3}}=\cfrac{\sqrt{3}}{3}
You can also use the table to look up the exact value.
3. What is the exact value of \sin (120)?
To find the exact value of the \sin 120, first find the reference angle to 120^{\circ}. \, 120^{\circ} is in the Quadrant II so to find the reference angle subtract 120^{\circ} from 180^{\circ}.
180-120=60
The reference angle is 60^{\circ}.
So, find the exact value of \sin(60) which you can find from the unit circle or the table.
The exact value of the \sin(60) is the same as the y -coordinate of the point at 60^{\circ} which is \cfrac{\sqrt{3}}{2}. Since 120^{\circ} is in Quadrant II, and the sine is positive in Quadrant II the exact value of \sin (120)=\cfrac{\sqrt{3}}{2}
You can find the value from the table too.
4. Find the exact value of \sec(45).
To find the exact value of \sec (45), you can use the unit circle or the table. 45^{\circ} is a Quadrant I angle and a reference angle and the secant is the reciprocal of the cosine. So, first find the x -coordinate of the point at 45^{\circ} on the unit circle.
The x -coordinate at 45^{\circ} is \cfrac{\sqrt{2}}{2}.
The reciprocal of \cfrac{\sqrt{2}}{2} is \cfrac{2}{\sqrt{2}} or \sqrt{2}, which means that the \sec(45)=\sqrt{2}
You can also get the values from the table.
5. What is the exact value of \cos(0)+\sin(30)?
To find the exact value of \cos (0)+\sin (30), first find the exact value of each trig function from the unit circle or using the table.
From the unit circle, \cos(0)=1 and \sin(30)=\cfrac{1}{2}
So, \cos(0)+sin(30)=1+\cfrac{1}{2}=1\cfrac{1}{2}
You can get the values from the table too.
6. What is the exact value of \cos(45)+\sin(60)?
To find the exact value of \cos (45)+\sin (60), you can use the unit circle or the table to find the exact value of each trig function.
The exact value of \cos (45)=\cfrac{\sqrt{2}}{2} and the exact value of \sin (60)=\cfrac{\sqrt{3}}{2} so,
\cos(45)+sin(60)=\cfrac{\sqrt{3}}{2}+\cfrac{\sqrt{2}}{2}=\cfrac{\sqrt{3}+\sqrt{2}}{2}
You can use the table too to find the exact values.
How to find the exact value of trig functions FAQs
The cofunction identities are relationships between trigonometric functions of complementary angles. For example, \sin\theta=\cos\left(90^{\circ}-\theta\right) and \cos\theta=\sin\left(90^{\circ}-\theta\right).
Yes, using the various angle identities and even half-angle and double angle identities can be helpful when finding exact values of trigonometric functions.
The next lessons are
- Circle math
- Angles of a circle
- Circle theorems
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