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.
