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

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

20:18

Video Transcript

Evaluating Trigonometric Function Values with Angles 30, 45, and 60

In this video, we will learn how to evaluate the trigonometric functions at angles of 30, 45, and 60 degrees. We’ll find these results geometrically by constructing right triangles. And we’ll also see how we can reverse this process by considering the question “What if we had two sides of a right triangle? Can we determine the angles of the right triangle?”

To do this, let’s start by recalling how we define the trigonometric functions for an angle 𝜃. We can recall that the trigonometric functions are defined based on the ratios of side lengths in a right triangle. To find the trigonometric functions evaluated at an angle of 𝜃, we start by sketching a right triangle with angle 𝜃. We then label the sides of this right triangle based on their position relative to angle 𝜃. We have the hypotenuse of the right triangle is the longest side in the right triangle. That’s the one opposite the right angle. Then, the side opposite angle 𝜃 is called the opposite side. Finally, the remaining side adjacent to angle 𝜃 is called the adjacent side.

This then allows us to define the trigonometric functions at angle 𝜃. First, the sin of 𝜃 will be the length of the opposite side to angle 𝜃 divided by the length of the hypotenuse. Next, the cos of angle 𝜃 will be the length of the adjacent side to angle 𝜃 divided by the length of the hypotenuse. Finally, the tan of angle 𝜃 is the length of the opposite side to angle 𝜃 divided by the length of the adjacent side to angle 𝜃.

Therefore, if we can construct a right triangle where we know the lengths of the right triangle and the internal angles of the right triangle, we can evaluate the trigonometric functions at this angle by using the right triangle. And there’s many different ways we can construct right triangles where we know the internal angles and the side lengths. Let’s go through two of these.

First, consider a square of unit length. We can then construct two right triangles by cutting the square along its diagonal. In fact, these two right triangles will be congruent, for example, by using the side-angle-side criterion. To use one of these right triangles to evaluate trigonometric functions, we’re going to need to know all three of its side lengths and its internal angles.

Let’s start by finding the internal angles. To do this, we can start by noting that the right triangles are isosceles. And in particular this tells us that two non-right angles of this right triangle are going to have the same measure. If we call this 𝜃, then we know since the sum of the internal angles in a triangle are 180 degrees and the right angle has measure 90 degrees, we know 𝜃 plus 𝜃 plus 90 degrees is 180 degrees. We can then solve this equation for 𝜃. Subtracting 90 degrees from both sides of the equation, we get that two 𝜃 is equal to 90 degrees. And then we divide through by two to get that 𝜃 is 45 degrees. And of course this makes sense. We’re cutting the square in half, so we can think of this as cutting the angle in half.

Let’s now find the missing length of this right triangle. We need to find the length of the hypotenuse given the length of the two other sides. We can do this by using the Pythagorean theorem. The Pythagorean theorem tells us, in any right triangle, the square of the hypotenuse is equal to the sum of the square of the two shorter sides in the right triangle. So if we call the length of the hypotenuse ℎ, we have ℎ squared is equal to one squared plus one squared. We can then solve this equation for ℎ. One squared plus one squared is equal to two. So we have ℎ squared is equal to two. And then we take the square root of both sides of the equation. Remember, ℎ is a length, so it must be positive. We get that ℎ is equal to the square root of two.

We can then add this to our diagram. And now we can see that we have a right triangle where we know all of the lengths and the internal angles. So we’re almost ready to use this right triangle to evaluate our trigonometric functions. However, remember, we still need to label the sides of this right triangle based on their position relative to our angle. First, as we’ve already said, the side with length root two will be the hypotenuse in this right triangle, since it’s the longest side opposite the right angle. Next, the side opposite the angle of 45 degrees is the opposite side. Finally, the remaining side adjacent to our marked angle of 45 degrees is the adjacent side.

We can now use this right triangle to evaluate the trigonometric functions at these values. We just need to substitute 𝜃 is 45 degrees, length of the adjacent side is one, the length of the opposite side is one, and the length of the hypotenuse is root two. Let’s start with the sine function. We get that sin of 45 degrees is equal to one divided by the square root of two. We could leave our answer like this. However, we can simplify this by rationalizing the denominator. We multiply both the numerator and denominator by root two to get root two divided by two. Therefore, we’ve shown the sin of 45 degrees is root two over two.

We can follow the same process for the cosine function. We get the cos of 45 degrees is one over root two, which is exactly the same value as we had above. So we can rationalize the denominator in exactly the same way to show that the cos of 45 degrees is also root two over two. Finally, by using the values from this right triangle, we can show the tan of 45 degrees is equal to one divided by one, which of course simplifies to give us one. Therefore, by using a unit square, the Pythagorean theorem, and the definition of the trigonometric functions, we were able to evaluate the sin of 45 degrees, the cos of 45 degrees, and the tan of 45 degrees.

And now before we use these to answer questions involving evaluating trigonometric expressions, there’s one more triangle we can use to evaluate the trigonometric functions at two different angles. This time, instead of starting with a unit square, we’re going to start with an equilateral triangle. Remember, the internal angles in an equilateral triangle are all 60 degrees. And it’s worth noting here we can choose any side length we want for our equilateral triangle. For example, we could use side length one. However, in this case, the arithmetic comes out easier if we use side length two. So we’re going to choose to use side length two. However, we could use any side length we want.

We can then construct two right triangles from our equilateral triangle by splitting our right triangle in two down the median line. Since this is a median line, we can find the base of our right triangle. It will have length one, since it splits the length of two in half. We can also find the missing internal angle in this right triangle since the sum of the internal angles in a right triangle add to 180 degrees. And we can see that 60 degrees plus 30 degrees plus 90 degrees is 180 degrees. So the missing angle has measure 30 degrees.

And finally we can find the missing side length of this right triangle 𝑙 by using the Pythagorean theorem. The square of the length of the hypotenuse is equal to the sum of the squares of the two shorter sides. So two squared is equal to 𝑙 squared plus one squared. We can then solve this equation for 𝑙. First, we have two squared is four and one squared is one. Then, we can subtract one from both sides of the equation to get that three is equal to 𝑙 squared. Finally, we take the square root of both sides of the equation. Well, we know that 𝑙 is a length, so it’s positive. This gives us that 𝑙 is equal to the square root of three. We can then add this to our diagram.

And now we see we have a right triangle where we know all of the side lengths of this right triangle and all of its internal angles. So, once again, we can use this right triangle to evaluate our trigonometric functions. To do this, we need to label the sides of this right triangle based on their position relative to the angle. However, this time, we have two choices. We can label the sides based on their position relative to 60 degrees or based on their position relative to 30 degrees. This will allow us to evaluate the trigonometric functions at 60 degrees and 30 degrees.

Let’s start by labeling the sides of this triangle based on their position relative to the angle of 60 degrees. We’ll do this by resketching the right triangle. First, as we’ve already discussed, the hypotenuse of this right triangle is two, since it’s the longest side opposite the right angle. Next, the side opposite the angle of 60 degrees is the opposite side. That’s the side of length root three. Finally, the remaining side adjacent to our angle of 60 degrees is the adjacent side. That’s the side of length one.

We can now use this to evaluate the sin, cos, and tan of 60 degrees. Let’s start with the sin of 60 degrees. It’s the length of the opposite side to 60 degrees divided by the hypotenuse. That’s root three divided by two. Next, the cos of 60 degrees is the length of the adjacent side divided by the length of the hypotenuse. This is one divided by two, or one-half. Finally, the tan of 60 degrees is the length of the opposite side divided by the length of the adjacent side. We can see this is root three divided by one, which simplifies to give us root three.

So this allowed us to evaluate the trigonometric functions at 60 degrees. Let’s now relabel our right triangle based on their position relative to 30 degrees to evaluate the trigonometric functions at 30 degrees. Once again, we’ll do this by resketching the right triangle. The hypotenuse stays the same; it’s still the side opposite the right angle. However, the sides labeled the opposite and adjacent sides switch around, since now the side of length one is opposite the angle of 30 degrees and the side of length root three is adjacent the angle of 30 degrees.

We can now use this right triangle to evaluate the sin, cos, and tan of 30 degrees. The sin of 30 degrees is the opposite divided by the hypotenuse. That’s one-half. The cos of 30 degrees is the adjacent divided by the hypotenuse. That’s root three over two. And the tan of 30 degrees is the opposite divided by the adjacent. That’s one divided by root three. And we can simplify this by rationalizing the denominator. We multiply both the numerator and denominator by root three to get root three over three. And this means we’ve shown how to evaluate the three trigonometric functions at 30 degrees, 45 degrees, and 60 degrees.

We can then clear some space and construct a table, giving us the results we’ve just shown. In the columns of this table, we have the angles in degrees. And in the rows of the table, we have the trigonometric functions. Then, the entry in this row and column tells us the trigonometric function evaluated at this angle. We can then use this table to evaluate the trigonometric functions at these three angles. For example, if we wanted to determine the cos of 60 degrees, we find the row of the table with the cosine function and the column with 60 degrees. Then, the entry in this table in this row and column tells us this value. The cos of 60 degrees is one-half.

And a table like this is very useful to commit to memory, so we don’t need to use the geometric construction every single time. However, it can still be useful to remember the geometric construction to be able to prove these results. Let’s now see an example where we’re asked to evaluate a trigonometric function.

Find the exact value of the sin of 30 degrees.

In this question, we’re asked to determine the exact value of a trigonometric function. We can see that the argument of this trigonometric function is 30 degrees. And there’s actually several different ways we can answer this question. For example, we could type the sin of 30 degrees into our calculator, and we would get an exact answer. However, it’s possible to answer this question without using a calculator, so let’s do that instead.

And there are two ways we can evaluate this expression without using a calculator. First, we can recall that 30 degrees is one of our special angles. We should commit all of the trigonometric functions to angles 30 degrees, 45 degrees, and 60 degrees to memory. One way of doing this is to use the following table. In the columns of our table, we have the angles 30 degrees, 45 degrees, and 60 degrees. And in the rows of our table, we have the three trigonometric functions.

We can then remember the row involving the sine function by recalling the sin of 30 degrees is root one over two, the sin of 45 degrees is root two over two, and the sin of 60 degrees is root three over two. The numerator is the square root of one, then the square root of two, then the square root of three. The row for the cosine function is then this in reverse. The last entry is root one over two. Then, the second entry is root two over two. And then the first entry is root three over two.

And we know the tan of 𝜃 is equal to the sin of 𝜃 divided by the cos of 𝜃. So we can find the row for tangent in our table by dividing the row for sine in the table by the row for cosine. In any case, we can use this table to evaluate the sin of 30 degrees. We need to find the entry in the column of the table for 30 degrees and the row of the table for the sin of 𝜃. And we can see that this entry is one-half. This then allows us to conclude the sin of 30 degrees is one-half.

And we could stop here. However, it can be difficult to memorize this table. So we can also briefly go over a geometric way of evaluating the sin of 30 degrees. First, we recall the trigonometric functions are the ratios of side lengths in a right triangle. So we can use a known right triangle to determine these trigonometric functions. In this case, we’ll construct a right triangle by using an equilateral triangle of side length two.

Remember, in an equilateral triangle, the internal angles are 60 degrees. We can split this equilateral triangle in two by using its perpendicular bisector. This cuts the base of the triangle in half, so the length of the base of this right triangle is one. The sum of the measures of the internal angles of a right triangle are 180 degrees. So the missing angle in this right triangle is 30 degrees.

Finally, we can determine the length of the missing side by using the Pythagorean theorem. The Pythagorean theorem tells us, in a right triangle, the square of the hypotenuse is equal to the sum of the square of the two shorter sides. So if we say the missing side has a length 𝑙, we have 𝑙 squared plus one squared is equal to two squared. We can solve this for 𝑙. Rearranging our equation, we have 𝑙 squared is equal to three. We then take the square root of both sides of the equation, where we remember 𝑙 is a length, so it’s positive. 𝑙 is the square root of three.

Now, either by recalling the definition of the sine function or by using the acronym SOHCAHTOA, we know that the sin of 𝜃 is equal to the length of the side opposite angle 𝜃 divided by the length of the hypotenuse. We want to do this for the angle 𝜃 is equal to 30 degrees. First, we can see that the hypotenuse is the side opposite the right angle. That’s the side of length two. Second, we can see that the side opposite the angle of 30 degrees is the side of length one. Substituting these values into our definition of the sine function, we get the sin of 30 degrees is one divided by two.

Therefore, we were able to find the exact value of the sin of 30 degrees in two different ways. In both ways, we showed the sin of 30 degrees is one-half.

In our next example, we’ll find the exact value of a trigonometric expression involving trigonometric functions evaluated at 30 degrees and 45 degrees.

Find the value of two cos 45 degrees times the sin of 30 degrees.

In this question, we’re asked to evaluate a trigonometric expression. And we can see the arguments of all of our trigonometric functions are 45 degrees and 30 degrees. And these are two of our special angles. We should commit all of the trigonometric functions evaluated at angles of 30 degrees, 45 degrees, and 60 degrees to memory. One way of doing this is to use a table of values as shown. In the columns, we have the angles of 30 degrees, 45 degrees, and 60 degrees. And in our rows, we have the trigonometric functions. In particular, the first row of this table will be one over root two, root two over two, and root three over two. And the second row of this table is the same as the first row of the table in reverse: root three over two, root two over two, root one over two.

We can then use this table to evaluate the cos of 45 degrees. We can see the entry in the same row is cos 𝜃 and the same column of 45 degrees is root two over two. So this tells us the cos of 45 degrees is root two over two. We can do the same for the sin of 30 degrees. We see that it’s equal to one-half. We now just substitute these values into our expression. This gives us two cos 45 degrees sin 30 degrees is equal to two times root two over two multiplied by one-half. We can then cancel the shared factor of two to get our final answer of root two divided by two.

So far, we’ve used right triangles to help us evaluate the trigonometric functions. However, it’s also possible to do this process in reverse. If we know the ratio of the side lengths of a right triangle, we can use this information to determine the angle of the right triangle. For example, suppose we were told the length of the opposite side divided by the length of the adjacent side was one. Then, there’s many different ways we can find the angle 𝜃.

One way is to note that the length of the opposite side is equal to the adjacent side, so we have an isosceles right triangle. This means the other unknown angle is also equal to 𝜃. So the measure of angle 𝜃 is 45 degrees. Adding this to our diagram, we can notice something interesting about this right triangle. This right triangle is similar to our right triangle with lengths one, one, and root two by the angle-angle-angle criterion. This is the right triangle which allowed us to determine the tan of 45 degrees was equal to one. So, a second way of answering this question is to note that these two right triangles are similar and then to scale this right triangle down to the given right triangle. Then, we can just use tan of 45 degrees is equal to one to conclude the angle has measure 45 degrees.

To help us better understand this relationship, let’s introduce the inverse trigonometric function and some of their properties. First, if 𝑎 is greater than zero and less than one, then 𝜃 is equal to the inverse sin of 𝑎 is the unique acute angle solution to the equation sin 𝜃 is equal to 𝑎 and 𝜃 is equal to the inverse cos of 𝑎 is the unique acute angle solution to the equation cos of 𝜃 is equal to 𝑎.

Next, if 𝑎 is positive, then 𝜃 is equal to the inverse tan of 𝑎 is the unique acute angle solution to the equation tan to 𝜃 is equal to 𝑎. The inverse trigonometric functions take as an input the ratio value of 𝑎, and they output 𝜃 which is the angle in the right triangle for this ratio. And in particular these properties tell us that the angle is unique. So we only need the ratio of the side lengths in the right triangle to determine the angle.

Let’s see an example of applying the inverse trigonometric functions to help us solve an equation.

If cos of 𝑥 is equal to one-half, find the value of 𝑥, where zero degrees is less than 𝑥 is less than 90 degrees.

In this question, we’re given a trigonometric equation involving 𝑥. We’re asked to find the value of 𝑥, and we’re told that 𝑥 is an acute angle. And since 𝑥 is an acute angle, we can say that 𝑥 is a possible angle in a right triangle. And then by using the acronym SOHCAHTOA, we can recall the cosine of an acute angle is the ratio of the length of the adjacent side to the angle divided by the length of the hypotenuse in the right triangle. So the equation cos of 𝑥 is equal to one-half is telling us the ratio of two side lengths in a right triangle. And then we can recall that we can solve this by using the inverse trigonometric functions. In particular, 𝜃 is equal to the inverse cos of 𝑎 is the unique acute angle solution to the equation cos 𝜃 is equal to 𝑎, where our value of 𝑎 must be between zero and one.

In this equation, we can see the value of 𝑎, the ratio, is one-half. Therefore, this property tells us 𝑥 is equal to the inverse cos of one-half. And we can evaluate the inverse cos of one-half by recalling the cos of 60 degrees is equal to one-half. And remember, our property tells us the acute angle solution to this equation is unique. So, because the cos of 60 degrees is equal to one-half, we can see that 𝑥 is equal to 60 degrees is a solution to our equation. Therefore, the inverse cos of one-half is equal to 60 degrees and 𝑥 is equal to 60 degrees is the solution to our equation.

Let’s now go over some of the key points of this video. First, we showed that we can evaluate trigonometric functions by constructing right triangles of known side lengths and known angles. In particular, we can use geometric results to construct the following two right triangles. We construct the first right triangle by splitting a square of unit length along its diagonal. And we construct the second right triangle by splitting an equilateral triangle of side length two along one of its median lines. Then, we can apply right triangle trigonometry to these two right triangles to evaluate the sine, cosine, and tangent functions at angles 30 degrees, 45 degrees, and 60 degrees. We can then construct a table of values to help us recall all of these evaluations.

Finally, we saw that we can use the inverse trigonometric functions to solve equations and to find missing angles. In particular, these inverse trigonometric functions give us unique acute angle solutions. So since we know the values of the sine, cosine, and tangent functions of 30 degrees, 45 degrees, and 60 degrees, we can use these values along with the fact that our solutions are unique up to acute angles to solve all of these equations.

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