Video Transcript
In this video, we will learn how to
derive the angle sum and difference identities graphically and use them to find
trigonometric values. The trigonometric angle sum and
difference identities have been used in mathematics for centuries to solve
real-world problems. The ancient Greeks used these
formulas to solve astronomy problems, like distances from Earth to the Sun. Before we try and derive some of
these identities or use them to solve problems, let’s remind ourselves what the sum
and difference identities are. For the angle sum identities, we
have sin of 𝐴 plus 𝐵 is equal to sin of 𝐴 times cos of 𝐵 plus cos of 𝐴 times
sin of 𝐵. The cos of 𝐴 plus 𝐵 is equal to
the cos of 𝐴 times the cos of 𝐵 minus the sin of 𝐴 times the sin of 𝐵. And finally, the tan of 𝐴 plus 𝐵
will be equal to the tan of 𝐴 plus the tan of 𝐵 all over one minus the tan of 𝐴
times the tan of 𝐵.
Note here that the variables 𝐴 and
𝐵 can represent any angle measure, and this will work when we’re operating in
radians or degrees. In some places, you might see these
angle measures represented with the variables 𝑥 and 𝑦 or even the Greek letters 𝛼
and 𝛽. We use the angle sum identity when
we want to take a larger angle and break it up into two smaller angles. Generally, we already know the sine
and cosine values for the two smaller angles, which help us calculate the sine,
cosine, or tangent of a larger angle measure. So we know sin of 𝐴 minus 𝐵 is
equal to sin of 𝐴 times cos of 𝐵 minus cos of 𝐴 times sin of 𝐵. cos of 𝐴 minus 𝐵 is equal to cos
of 𝐴 times cos of 𝐵 plus sin of 𝐴 times sin of 𝐵. And tan of 𝐴 minus 𝐵 is equal to
the tan of 𝐴 minus the tan of 𝐵 all over one plus the tan of 𝐴 times the tan of
𝐵.
Now, we’re gonna take this first
angle sum identity and form a geometric proof to show that it’s true. To prove the sin of 𝐴 plus 𝐵 is
equal to the sin of 𝐴 times the cos of 𝐵 plus the cos of 𝐴 times the sin of 𝐵,
we can use what we know about right-angled triangles. Suppose we have a right-angled
triangle and we know that one of these angles measures 𝐵 degrees. Stacked on top of that is the right
triangle 𝐷𝐶𝑋, which has an angle measure of 𝐴 degrees. By drawing a perpendicular line
from 𝐷 to the line 𝑋𝑌, we’ve created a third right triangle, and this third right
triangle has an angle of 𝐴 plus 𝐵.
We can then use this figure to form
a geometric proof. Let’s let the distance from 𝑋 to
𝐷 be equal to one. In our figure, what is the sin of
𝐴 plus 𝐵? That would be inside triangle
𝐷𝐹𝑋. And we know that in a right-angled
triangle, the sin of an angle is equal to the opposite side length over the
hypotenuse. So the sin of 𝐴 plus 𝐵 will be
equal to the side length 𝐷𝐹 over one. In our figure, 𝐷𝐹 can be broken
up into two smaller pieces. If we label this point 𝐸, then we
can say that 𝐷𝐹 is equal to 𝐷𝐸 plus 𝐸𝐹.
Now we want to see if we can find
the sin or cos of angle 𝐴 and the sin or cos of angle 𝐵. If we look at the right triangle
that includes angle 𝐴, we see that it has a hypotenuse of one, which means the sin
of angle 𝐴 will be the opposite side length 𝐷𝐶 over one, which simplifies to
𝐷𝐶. And we can label that side length
the sin of 𝐴. Similarly, for cos of 𝐴, we know
that the cos of any angle is equal to the adjacent side length over the
hypotenuse. Here, that’s going to be the side
length 𝑋𝐶 over one, which simplifies to 𝑋𝐶. And so we can label that side
length cos of 𝐴.
Now we’ll consider the right-angled
triangle containing angle 𝐵. To find the sin of angle 𝐵, we see
the opposite side length, 𝐶𝑌, over the hypotenuse, which here is the cos of angle
𝐴. At this point, we recognize
something interesting. We have an equation in which we
have two terms, the cos of 𝐴 and the sin of 𝐵. We can rearrange this equation by
multiplying both sides by cos of 𝐴. And then we see that the line
segment 𝐶𝑌 is equal to the cos of 𝐴 times the sin of 𝐵. We already know that the sin of 𝐴
plus 𝐵 will be equal to line segment 𝐷𝐸 plus line segment 𝐸𝐹. And now we’re saying that the cos
of 𝐴 times the sin of 𝐵 is equal to the line segment 𝐶𝑌. But we want to ask the question,
can we relate line segment 𝐸𝐹 to line segment 𝐶𝑌?
If we look at our figure, line
segment 𝐶𝑌 and line segment 𝐸𝐹 fall inside a rectangle, and they are opposite
sides from each other. This means line segment 𝐶𝑌 is
equal in length to line segment 𝐸𝐹. And we can substitute that value in
for the cos of 𝐴 times the sin of 𝐵. Now we have line segment 𝐸𝐹 on
both sides of this equation. And that means we want to ask the
question, can we relate the sin of 𝐴 times the cos of 𝐵 to the line segment
𝐷𝐸? Line segment 𝐷𝐸 is here. And if we notice that line segment
𝐸𝐶 and line segment 𝑋𝑌 are parallel, they’re cut by the transversal 𝑋𝐶, which
means we can say that this angle is equal to the angle 𝐵. And then we can say that this angle
is 90 minus 𝐵. Therefore, this angle is also angle
𝐵.
Using this angle as our angle 𝐵,
we can say that the cos of angle 𝐵 equals the adjacent side length 𝐷𝐸 over its
hypotenuse, which is the sin of 𝐴. So we have cos of 𝐵 is equal to
𝐷𝐸 over sin of 𝐴. And if we multiply both sides of
this equation by the sin of 𝐴, we get that line segment 𝐷𝐸 is equal to the sin of
𝐴 times the cos of 𝐵. If we substitute line segment 𝐷𝐸
in for the sin of 𝐴 times the cos of 𝐵, we’ve shown that, in this figure, the sin
of 𝐴 plus 𝐵 is equal to line segment 𝐷𝐸 plus line segment 𝐸𝐹. The sin of 𝐴 times the cos of 𝐵
is equal to line segment 𝐷𝐸. The cos of 𝐴 times the sin of 𝐵
is equal to line segment 𝐸𝐹. And therefore, line segment 𝐷𝐸
plus line segment 𝐸𝐹 is equal to line segment 𝐷𝐸 plus line segment 𝐸𝐹.
It would also be possible to use
this figure to derive the cos of 𝐴 plus 𝐵 and then to find the tan of 𝐴 plus
𝐵. But you can do that another
day. For now, let’s look at some
examples of using the angle sum and difference identities. To solve problems with the angle
sum and difference identities, we’ll often have to recognize patterns and/or
rearrange equations. So let’s look at our first example
of that.
Simplify cos of two 𝑋 times cos of
22𝑋 minus sin of two 𝑋 times sin of 22𝑋.
When we look at this expression, we
see that we have the angle two 𝑋 and the angle 22𝑋. We have the cos and sin of the
angle two 𝑋, and we have the cos and sin of the angle 22𝑋, which means we have an
expression in the form cos of 𝐴 times cos of 𝐵 minus sin of 𝐴 times sin of
𝐵. And we can simplify this expression
using the angle sum identities. We know that the cos of 𝐴 times
the cos of 𝐵 minus the sin of 𝐴 times the sin of 𝐵 is equal to the cos of 𝐴 plus
𝐵. In our case, the angle 𝐴 is two 𝑋
and the angle 𝐵 is 22𝑋, making this expression equal to cos of two 𝑋 plus 22𝑋,
cos of 24𝑋. And so the simplified form of this
expression is cos of 24𝑋.
Now we’re ready to consider another
example.
Given that sin of 60 degrees times
cos of 30 degrees minus cos of 60 degrees times sin of 30 degrees equals sin of 𝜃
degrees, find the value of 𝜃 in degrees.
When we look at this equation, we
notice that we’re dealing with angles of 60 degrees and two angles of 30
degrees. And this should remind us of an
angle sum or difference identity. That is sin of 𝐴 times cos of 𝐵
minus cos of 𝐴 times sin of 𝐵 is equal to sin of 𝐴 minus 𝐵. If we let 𝐴 be equal to 60 degrees
and 𝐵 be equal to 30 degrees, we can write the angle 𝜃 in terms of 𝐴 and 𝐵. 𝜃 is going to be equal to 𝐴 minus
𝐵. 𝜃 will then be equal to 60 degrees
minus 30 degrees, which equals 30 degrees. Using the angle difference
identities, we’re able to show that 𝜃 here is equal to 30 degrees.
In our next example, we’re given
the cosine values of two smaller angles and asked to find their cosine value
combined.
Find the cos of 𝐴 plus 𝐵 given
the cos of 𝐴 equals 15 over 17 and the cos of 𝐵 equals five over 13, where 𝐴 and
𝐵 are acute angles.
When we see this cos of 𝐴 plus 𝐵,
it should remind us of our angle sum identities. We know that the cos of 𝐴 plus 𝐵
is equal to the cos of 𝐴 times the cos of 𝐵 minus the sin of 𝐴 times the sin of
𝐵. We have enough information for the
first term since we’ve already been given the value of cos of 𝐴 and cos of 𝐵. So how should we go about finding
the sin of 𝐴 and the sin of 𝐵 if we know the cosine ratios? They are acute angles. And therefore, one strategy to find
sin 𝐴 and sin 𝐵 would be to make right-angled triangles with these
proportions.
First, we can draw a right triangle
with angle 𝐴. We know that the cosine
relationship will be equal to the adjacent side length over the hypotenuse. And so we would label the adjacent
side to 𝐴 15 and the hypotenuse 17. To find the sine relationship,
we’ll need to know this opposite side length, and that means we’ll need to use the
Pythagorean theorem. We’ll let our unknown side be
lowercase 𝑎. And then we’ll have 17 squared
equals 𝑎 squared plus 15 squared, which will give us 289 is equal to 𝑎 squared
plus 225. To isolate 𝑎, we subtract 225 from
both sides, and then we get 64 is equal to 𝑎 squared. Taking the square root of both
sides, we get 𝑎 equal to eight.
We’re only interested in the
positive square root since we’re dealing with distance. If 𝑎 equals eight, then the sin of
angle 𝐴 will be equal to eight over 17. If we consider a second
right-angled triangle with angle 𝐵, its adjacent side length is five and its
hypotenuse is 13. We should recognize that this is a
Pythagorean triple. It’s a set of positive integers
that occur in the ratio 𝑎 squared plus 𝑏 squared equals 𝑐 squared. And because we know the hypotenuse
is 13 and one of the sides is five, this is a five, 12, 13 triangle. And therefore, the adjacent side
length will be 12 and the sin of angle 𝐵 is 12 over 13. And we can plug that value in. 15 over 17 times five over 13 is 75
over 221. Eight over 17 times 12 over 13 is
96 over 221, which makes the cos of 𝐴 plus 𝐵 equal to negative 21 over 221.
Before we finish, we’ll look at one
final example involving right triangles and combining two different angles.
The diagram shows triangle
𝐴𝐵𝐶. Given that 𝐴𝐷 is perpendicular to
𝐵𝐶, 𝐴𝐷 equals 15 centimeters, 𝐵𝐷 equals 10 centimeters, and 𝐶𝐷 equals seven
centimeters, find the value of the tan of 𝑥 plus 𝑦.
The first thing we want to do is
label our diagram. 𝐴𝐷 equals 15 centimeters, 𝐵𝐷
equals 10 centimeters, and 𝐶𝐷 equals seven centimeters. We’re interested in the tangent of
this angle, the angle 𝑥 plus 𝑦. However, this angle is in a
triangle that’s not a right triangle, so we’ll need a different strategy to solve
for this angle. If we look, we see that angle 𝑥
and angle 𝑦 are both located inside right triangles. This means it’s possible to find
the tangent value of 𝑥 and the tangent value of 𝑦. And by our angle sum identity, we
know that the tan of 𝐴 plus 𝐵 will be equal to the tan of 𝐴 plus the tan of 𝐵
all over one minus the tan of 𝐴 times the tan of 𝐵.
The tangent of any angle in a right
triangle will be equal to the opposite side length over the adjacent side
length. For angle 𝑥, the opposite side
length is 10 and the adjacent side length is 15. The tan of 𝑥 equals 10 over
15. To find the tan of 𝑦, we have the
opposite side length of seven over the adjacent side length of 15. So the tan of 𝑦 equals seven
fifteenths. So the tan of 𝑥 plus 𝑦 is equal
to 10 over 15 plus seven over 15 all over one minus 10 over 15 times seven over
15. Our numerator becomes 17 over
15.
Before we do this multiplying in
the denominator, we can do some simplifying. 10 over 15 simplifies to
two-thirds, and two-thirds times seven fifteenths equals 14 over 45. In our denominator, since we have
one minus 14 over 45, we can rewrite the one as 45 over 45. 45 minus 14 is 31. This means we’re dividing seventeen
fifteenths by 31 over 45. And to divide by a fraction, we
multiply by the reciprocal. We have seventeen fifteenths times
45 over 31. 15 goes into 45 three times. 17 times three equals 51, making
the tan of 𝑥 plus 𝑦 51 over 31.
Before we finish, let’s review the
key points from this video. Angle sum and difference identities
can be used to simplify expressions involving two angles. There are three angle sum
identities for sin of 𝐴 plus 𝐵, cos of 𝐴 plus 𝐵, and tan of 𝐴 plus 𝐵 and three
angle difference identities for sin of 𝐴 minus 𝐵, cos of 𝐴 minus 𝐵, and tan of
𝐴 minus 𝐵, all of which are listed here.
