Lesson Explainer: Evaluating Trigonometric Function Values with Angles 30, 45, and 60 Mathematics

In this explainer, we will learn how to find the trigonometric function values for 30-, 45-, and 60-degree angles.

We start by recalling what is meant by trigonometric functions. In a right triangle, we can label the lengths of the sides of the triangle relative to one of the internal angles. For example, we can label the sides in relation to angle πœƒ as shown in the diagram below.

We then define the trigonometric functions as the ratios of the lengths of the sides of the right triangle as follows: sinoppositehypotenusecosadjacenthypotenusetanoppositeadjacentπœƒ=,πœƒ=,πœƒ=.

Since we can determine the values of trigonometric functions from the ratios of side lengths in right triangles, we will construct a few right triangles geometrically to allow us to evaluate the trigonometric functions at specific values. We will start by considering a square of side length 2.

It is worth noting that we can choose any side length for the square. We chose 2 since it will make the arithmetic easiest. We split the square into two congruent right triangles along one of its diagonals as shown.

Since these are isosceles right triangles, the non-right angles must be equal. In particular, since the internal angles in a triangle sum to 180∘, we can determine that the missing angles are both 180βˆ’902=45∘∘∘.

We can then apply the Pythagorean theorem to find the length of the missing side. Recall that this tells us that the square of the hypotenuse is equal to the sum of squares of the other two sides. This gives us 𝑐=2+2𝑐=4+4𝑐=8.

We can solve this equation by taking the square root of both sides of the equation and noting that lengths must be positive: 𝑐=√8=√4Γ—2=2√2.

We can then use this right triangle to determine the values of trigonometric functions evaluated at 45∘. Labeling the sides of the triangle in relation to their position with respect to the bottom left angle of 45∘, we get the following.

By applying right triangle trigonometry, we get the following: sinoppositehypotenusecosadjacenthypotenusetanoppositeadjacent45==22√2=1√2=1√2Γ—βˆš2√2=√22,45==22√2=√22,45==22=1.∘∘∘

We can use exactly the same technique to evaluate the trigonometric functions at other angles. For example, we could use an equilateral triangle of side length 2, which we recall has internal angles equal to 60∘.

We can then split this into two right triangles using a perpendicular bisector as shown.

We will consider only one of these right triangles. We note that since the internal angles in a triangle sum to 180∘, we can determine the missing angle to be 180βˆ’60βˆ’90=30∘∘∘∘. We can also note that half of the base of the triangle is of length 1, giving us the following.

Since this is a right triangle, we can find the length of the missing side using the Pythagorean theorem. Recall that this tells us the square of the hypotenuse is equal to the sum of squares of the other two sides. This gives us 2=1+π‘Ž4=1+π‘Žπ‘Ž=3.

We can find the value of π‘Ž by taking the square root of both sides of the equation and noting that π‘Ž is a length and so it must be positive: π‘Ž=√3.

We can add this to our diagram to get the following right triangle.

We can use this to determine the values of the trigonometric functions evaluated at 30∘ and 60∘. First, we will label the sides relative to the angle 60∘.

Since sinoppositehypotenuseπœƒ=, we have sin60=√32.∘

Similarly, since cosadjacenthypotenuseπœƒ= and tanoppositeadjacentπœƒ=, we have costan60=12,60=√31=√3.∘∘

We can also determine the values of these trigonometric functions at 30∘ by labeling the sides of the triangle in relation to their position with respect to the angle 30∘.

We have sinoppositehypotenusecosadjacenthypotenusetanoppositeadjacent30==12,30==√32,30==1√3=1√3Γ—βˆš3√3=√33.∘∘∘

We can summarize the results we have shown in the following table.

Standard Fact: Evaluating Trigonometric Function Values with Angles of 30Β°, 45Β°, and 60Β°

By considering the ratios of certain right triangles, we can construct the following table that tells us the values of the trigonometric functions: sine, cosine, and tangent evaluated at the angles of 30∘, 45∘, and 60∘.

πœƒβˆ˜
30∘45∘60∘
sinπœƒ12√22√32
cosπœƒβˆš32√2212
tanπœƒβˆš331√3

These trigonometric results are useful and they are either worth committing to memory or we should be comfortable with how they are found geometrically. Let’s now see some examples of using these values to simplify or evaluate trigonometric expressions.

Example 1: Finding the Values of Trigonometric Functions of the Special Angles

Find the exact value of sin30∘.

Answer

We can determine the exact value of sin30∘ by using a calculator; we would get that sin30=12.∘

However, we should be comfortable with finding the values of the three trigonometric functions: sine, cosine, and tangent evaluated at 30∘, 45∘, and 60∘. To do this, we start by constructing an equilateral triangle with side length 2.

We then construct the following right triangle by considering the perpendicular bisector of the base.

Since the internal angles in a triangle sum to 180∘, we can find the missing angle to be 180βˆ’90βˆ’60=30∘∘∘∘. We can determine the length of the missing side by applying the Pythagorean theorem, which states that in a right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides. This gives 2=1+π‘Ž, which we can solve for π‘Ž by noting this value must be positive: π‘Ž=4βˆ’1π‘Ž=3π‘Ž=√3.

This gives us the following right triangle.

Since we want to determine the value of sin30∘, we will label the sides by referencing their position relative to the angle 30∘.

We recall that the sine function is the ratio of the length of the opposite side and the hypotenuse in a right triangle, giving us sinoppositehypotenuse30==12.∘

Hence, sin30=12∘.

In the previous example, we used right triangles to determine the trigonometric ratios at a specific angle. In practice, this is quite time consuming, so it is easier to just commit the table of values to memory and then apply these values to evaluate the expressions. In the remaining examples, we will just refer to the table.

Example 2: Using Trigonometric Values of Special Angles to Evaluate Trigonometric Expressions

Find the value of 24530cossin∘∘.

Answer

We can determine this value by using a calculator. However, it is a useful skill to be able to evaluate the trigonometric functions at angles of 30∘, 45∘, and 60∘ without a calculator, so we will answer this question without a calculator.

We can recall the full table of values for trigonometric functions at angles of 30∘, 45∘, and 60∘, as follows.

πœƒβˆ˜
30∘45∘60∘
sinπœƒ12√22√32
cosπœƒβˆš32√2212
tanπœƒβˆš331√3

Or alternatively, we can just recall that sin30=12∘ and cos45=√22∘. We then substitute these values into the expression to get 24530=2Γ—βˆš22Γ—12=√22.cossin∘∘

Hence, 24530=√22cossin∘∘.

Example 3: Using Trigonometric Values of Special Angles to Evaluate Trigonometric Expressions

Find the value of cossinsintantan6030βˆ’6060+30∘∘∘∘∘ without using a calculator.

Answer

We can evaluate this expression exactly, without using a calculator, by using geometry and right triangle trigonometry. However, it is easier to recall the following table of values.

πœƒβˆ˜
30∘45∘60∘
sinπœƒ12√22√32
cosπœƒβˆš32√2212
tanπœƒβˆš331√3

We see that sin30=12∘, cos60=12∘, tan30=√33∘, and tan60=√3∘. We can then substitute these values into the expression to evaluate. We have cossinsintantancossinsintantan6030βˆ’6060+30=6030βˆ’6060+(30)=ο€Ό12οˆο€Ό12οˆβˆ’ο€Ώβˆš32ο‹ο€»βˆš3+ο€Ώ1√3=14βˆ’32+13.∘∘∘∘∘∘∘∘∘∘

We then make sure the fractions have a common denominator so we can add them: cossinsintantan6030βˆ’6060+30=14βˆ’32+13=14Γ—33βˆ’32Γ—66+13Γ—44=312βˆ’1812+412=βˆ’1112.∘∘∘∘∘

Hence, the exact value of the trigonometric expression is βˆ’1112.

We have seen how we can use known triangles to help us evaluate trigonometric functions at certain angles. We can use this same logic to go in reverse; if the ratio of side lengths in a right triangle is a value we know, then we can determine the angle.

For example, imagine we have the following triangle.

And we are told that the ratio of the opposite and adjacent sides is 1; in other words, tanπœƒ=1. Then, there are two ways we can determine angle πœƒ. First, we could do this geometrically. We know oppositeadjacent=1, so oppositeadjacent=.

This means that this is an isosceles right triangle. Hence, the other unknown angle is also πœƒ.

We then use the fact that the sum of the internal angles in a triangle is 180∘ to find πœƒ: πœƒ+πœƒ+90=1802πœƒ=90πœƒ=45.∘∘∘∘

This is not the only method we can use to determine the value of πœƒ. We know that tan45=1∘ and that 45∘ is a possible angle in a right triangle since it is acute. So, we can just conclude that πœƒ is 45∘ using our knowledge that tan45=1∘.

We can also solve this example by using a fact about inverse trigonometric functions.

Definition: Inverse Trigonometric Functions for Acute Angles

For 0<π‘Ž<1,

  • πœƒ=π‘Žsin is the unique acute angle solution to the equation sinπœƒ=π‘Ž;
  • πœƒ=π‘Žcos is the unique acute angle solution to the equation cosπœƒ=π‘Ž.

For π‘Ž>0,

  • πœƒ=π‘Žtan is the unique acute angle solution to the equation tanπœƒ=π‘Ž.

Therefore, because tan45=1∘, we know πœƒ=45∘ is a solution to the equation tanπœƒ=1, and we know the solution is unique for acute angles. We know that tan∘1=45. We can do the same for all of the other values we know. For example, since sin60=√32∘, we know that sinοŠ±οŠ§βˆ˜ο€Ώβˆš32=60. We can write all of these in the following table.

Standard Fact: Evaluating the Inverse Trigonometric Functions at Certain Values

The following table tells us the values of the inverse trigonometric functions evaluated at the given values.

π‘Ž
12√22√32
sinοŠ±οŠ§π‘Ž30∘45∘60∘
cosοŠ±οŠ§π‘Ž60∘45∘30∘
π‘Ž
√331√3
tanοŠ±οŠ§π‘Ž30∘45∘60∘

Let’s now see an example of solving a trigonometric equation.

Example 4: Using an Inverse Trigonometric Function to Find a Special Angle

If cos(π‘₯)=12, find the value of π‘₯, where 0<π‘₯<90∘∘.

Answer

We first note that π‘₯ is an acute angle and we can solve trigonometric equations by using the inverse trigonometric functions. Since cos(π‘₯)=12, we take the inverse cosine of both sides of the equation to see that π‘₯=ο€Ό12.cos

Since π‘₯ is acute, we know it is an angle in a right triangle. We recall that cos60=12∘, so we can conclude that cosοŠ±οŠ§βˆ˜ο€Ό12=60. This is in the interval given, so π‘₯=60∘.

In our final example, we will rearrange and solve a trigonometric equation using our knowledge of the inverse trigonometric functions.

Example 5: Using an Inverse Trigonometric Function to Find a Special Angle

Given that √3(π‘₯)+2=3tan, find the value of π‘₯, where 0<π‘₯<90∘∘.

Answer

To solve a trigonometric equation, we should start by trying to simplify the equation; we have √3(π‘₯)+2=3√3(π‘₯)=1(π‘₯)=√33.tantantan

We then recall that since we are looking for acute solutions, π‘₯=ο€Ώβˆš33tan is the unique solution to this equation.

However, we know how to evaluate trigonometric functions at special values. In particular, we know that tan(30)=√33∘. Since the acute solution is unique, we must have that π‘₯=30∘.

Let’s finish by recapping some of the important points of this explainer.

Key Points

  • We can evaluate the trigonometric functions by constructing right triangles and using the ratios of the lengths of the sides.
  • In particular, we can construct the following two right triangles by using half a square and half an equilateral triangle of length 2.
  • By applying right triangle trigonometry to these triangles, we can evaluate the sine, cosine, and tangent functions at 30∘, 45∘, and 60∘. We get the following table of values.
    πœƒβˆ˜
    30∘45∘60∘
    sinπœƒ12√22√32
    cosπœƒβˆš32√2212
    tanπœƒβˆš331√3
  • For 0<π‘Ž<1,
    • πœƒ=π‘Žsin is the unique acute angle solution to the equation sinπœƒ=π‘Ž;
    • πœƒ=π‘Žcos is the unique acute angle solution to the equation cosπœƒ=π‘Ž.
  • For π‘Ž>0,
    • πœƒ=π‘Žtan is the unique acute angle solution to the equation tanπœƒ=π‘Ž.
  • The following table tells us the values of the inverse trigonometric functions evaluated at the given values.
    π‘Ž
    12√22√32
    sinοŠ±οŠ§π‘Ž30∘45∘60∘
    cosοŠ±οŠ§π‘Ž60∘45∘30∘
    π‘Ž
    √331√3
    tanοŠ±οŠ§π‘Ž30∘45∘60∘

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