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Trigonometric Functions
Trigonometric functions are functions on angles of right-angled triangles.
Trigon meant triangle in Ancient Greece and a metric is a type of measurement (in case you are
wondering where that word came from).
The functions we will look at in this section are the sine (sin), cosine (cos) and tangent (tan)
functions.
Here’s how they work:
Given a right-angled triangle, with angle 𝑥
and sides hypotenuse, opposite and adjacent,
with lengths ℎ, 𝑜𝑝 and a respectively:
sin(𝑥) =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑜𝑝
=
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
ℎ
cos(𝑥) =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝑎
=
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
ℎ
tan(𝑥) =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑜𝑝
=
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝑎
A good way to remember these are to memorise the following words; SOH, CAH, TOA (where,
for example, S stands for sine, O stands for opposite and H stands for hypotenuse).
Notice that all of these three functions take in a real number, 𝑥 (an angle of a triangle), and spit
out another real number (a length divided by another length).
Now, you may have spotted a potential problem here. The only thing that goes into these
functions, is the angle 𝑥. The thing that comes out is a length divided by another length.
What happens if we have another right angled triangle, with the same angle 𝑥, but different
lengths?
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According to our definitions above:
𝑏
𝑒
= sin(𝑥) =
𝑎
𝑑
𝑐
𝑓
= cos(𝑥) =
𝑎
𝑑
𝑏
𝑒
= tan(𝑥) =
𝑐
𝑓
𝑒
𝑏
Hence, for our definitions to make sense, it must be that 𝑓 = 𝑐 ,
𝑓
𝑑
𝑐
𝑒
𝑏
= 𝑎 and 𝑑 = 𝑎. So we better
check that this is true. To do so, we’re going to be a little tricky and arrange our two triangles in
the following way.
Adding a few dotted lines gives the outline of a bigger right-angled triangle with corners A, B
and C, side lengths 𝑎 + 𝑑, 𝑒 + 𝑏 and 𝑐 + 𝑓.
1
Notice that the area of triange ABC = 2 × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 =
1
2
(𝑐 + 𝑓)(𝑒 + 𝑏).
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We can also calculate the area buy observing that the triangle ABC is made up of two smaller
triangles and a rectangle. Hence, its area is equal to the sum of these three areas:
1
Area of triangle ABC = 2 𝑐𝑑 + 𝑓𝑏 +
1
2
𝑒𝑓.
Thus, since these two areas must be the same (as they are the area of the same shape):
1
1
1
(𝑐 + 𝑓)(𝑒 + 𝑏) = 𝑐𝑏 + 𝑓𝑏 + 𝑒𝑓
2
2
2
⤇
1
2
𝑐𝑏 +
1
⤇
2
1
⤇
2
1
2
𝑐𝑒 +
𝑐𝑒 =
𝑒𝑓 +
1
2
1
2
𝑐𝑒 +
1
2
𝑓𝑏 =
1
2
𝑐𝑏 + 𝑓𝑏 +
1
2
2
(Expanding the left side.)
1
𝑓𝑏 = 𝑓𝑏
1
𝑒𝑓
1
𝑓𝑏
(Subtracting 2 𝑓𝑏 from both sides.)
⤇ 𝑐𝑒 = 𝑓𝑏
𝑒
⤇
𝑓
(Multiplying both sides by 2.)
𝑏
=
(Dividing both sides by f and c.)
𝑐
𝑏
Hence, 𝑐 = tan(𝑥) =
1
(Subtracting 2 𝑐𝑏 and 2 𝑒𝑓 from both sides.)
𝑒
𝑓
and the tangent function is properly defined.
For sine and cosine, observe that by Pythagorus, we know that 𝑓 = √𝑑 2 − 𝑒 2 and
𝑐 = √𝑎2 − 𝑏 2 . Thus:
𝑏
√𝑎2 − 𝑏 2
⤇ 𝑏 2 (𝑑 2 − 𝑒 2 ) = 𝑒 2 (𝑎2 − 𝑏 2 )
=
𝑒
√𝑑 2 − 𝑒 2
(Squaring both sides and cross multiplying.)
⤇ 𝑏 2 𝑑2 = 𝑒 2 𝑎2
(Adding 𝑏 2 𝑒 2 to both sides.)
⤇ 𝑏𝑑 = 𝑒𝑎
𝑏
⤇𝑎 =
(Square rooting both sides – how do we know to take the positive root?)
𝑒
𝑑
𝑏
Hence, 𝑎 = sin(𝑥) =
𝑒
𝑑
and the sine function is properly defined.
It is left as an exercise to show that the cosine function is properly defined.
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Everything looks good. Although there is another problem you may have spotted. What exactly
is our domain? What is the set of inputs?
We know that these functions eat real numbers and spit out real numbers. But exactly which
real numbers?
Our picture looks good for 0 < 𝑥 < 90 (degrees), but what do we do for 0 < 𝑥 < 180?
The answer is that we reflect the triangle about a vertical line, give the adjacent side a negative
length instead its positive one and measure the angle from the other side, as shown:
Then the calculations are then almost the same, except the length of the adjacent side is now
negative instead of positive.
Why the heck would our adjacent side all of a sudden have a negative length, though? That
must be confusing.
To find out, let use impose our triangles over our 𝑥 and 𝑦 axis:
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If a is the length of the adjacent side of the triangle, when we reflect it about the 𝑦 axis, its
value on the 𝑥 axis becomes –a. (Things may be a little confusing now that the 𝑥 axis is
introduced, but remember that 𝑥 is the value of the angle and is not related to the 𝑥 axis here.)
Notice that the length of the opposite side is still represented by op in the 𝑦 axis, even after the
reflection.
The length of the hypotenuse is still ℎ as it is unrepresented on the 𝑥 and 𝑦 axis.
Observe from the diagram and the triangle on the right that for 90 < 𝑥 < 180 we have:
sin(𝑥) =
𝑜𝑝
= sin(180 − 𝑥)
ℎ
cos(𝑥) =
−𝑎
𝑎
= − = −cos(180 − 𝑥)
ℎ
ℎ
tan(𝑥) =
𝑜𝑝
𝑜𝑝
= −
= −tan(180 − 𝑥)
−𝑎
𝑎
Note that 0 < 180 − 𝑥 < 90, so we can calculate the right hand sides by our original
definitions.
Hence, all three of our new functions are defined for real numbers 𝑥 such that 0 < 𝑥 < 90 and
90 < 𝑥 < 180. So far, these are the numbers included in the domains of these functions.
We can extend their domains even further by treating the cases where 180 < 𝑥 < 270 and
270 < 𝑥 < 360. We do this by flipping our triangles upside down, reflecting them in the 𝑥 axis:
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For 180 < 𝑥 < 270, measure the angle 𝑥 from the 𝑥 axis as usual to the edge of the triangle
reflected in both axis (shown in red in the diagram).
Notice that for this triangle (triangle 3. In the diagram below), the lengths of the adjacent side
and the opposite side are both negative.
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Hence, for 180 < 𝑥 < 270 we have the following results:
sin(𝑥) =
−𝑜𝑝
𝑜𝑝
= −
= −sin(𝑥 − 180)
ℎ
ℎ
cos(𝑥) =
tan(𝑥) =
−𝑎
𝑎
= − = −cos(𝑥 − 180)
ℎ
ℎ
−𝑜𝑝
𝑜𝑝
=
= tan(𝑥 − 180)
−𝑎
𝑎
Note once again that 0 < 𝑥 − 180 < 90, so we can calculate the right hand sides by our
original definitions.
It is left as an exercise to show that for 270 < 𝑥 < 360 we have:
sin(𝑥) =
−𝑜𝑝
𝑜𝑝
= −
= −sin(360 − 𝑥)
ℎ
ℎ
cos(𝑥) =
tan(𝑥) =
𝑎
= cos(360 − 𝑥)
ℎ
−𝑜𝑝
𝑜𝑝
= −
= −tan(360 − 𝑥)
𝑎
𝑎
Where 0 < 360 − 𝑥 < 90.
When our angle 𝑥 = 0, the triangle is pretty much just a line from 0 to a along the 𝑥 axis, where
a is the length of the adjacent side. So declare op = 0, where op is the length of the opposite
side, as before. Also, the hypotenuse of this “triangle” shares a length with its adjacent side.
Therefore, for 𝑥 = 0 we have:
sin(0) =
𝑜𝑝 0
= =0
ℎ
ℎ
cos(0) =
𝑎
𝑎
= =1
ℎ
𝑎
tan(0) =
𝑜𝑝
0
= = 0
𝑎
𝑎
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It is left as an exercise to convince yourself that the following definitions are sensible:
For 𝑥 = 90 we have:
sin(90) =
𝑜𝑝 𝑜𝑝
=
= 1
ℎ
𝑜𝑝
cos(90) =
𝑎
0
= =0
ℎ
ℎ
sin(180) =
𝑜𝑝 0
= =0
ℎ
ℎ
For 𝑥 = 180 we have:
−𝑎
−𝑎
=
= −1
ℎ
𝑎
cos(180) =
tan(180) =
𝑜𝑝
0
=
= 0
−𝑎
−𝑎
For 𝑥 = 270 we have:
sin(270) =
−𝑜𝑝 −𝑜𝑝
=
= −1
ℎ
𝑜𝑝
cos(270) =
𝑎
0
=
=0
ℎ
𝑜𝑝
For 𝑥 = 360 we have:
sin(360) =
𝑜𝑝 0
= = 0 = sin(0)
ℎ
ℎ
cos(360) =
𝑎
𝑎
= = 1 = cos(0)
ℎ
𝑎
tan(360) =
𝑜𝑝
0
= = 0 = tan(0)
𝑎
𝑎
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Did I miss something? What about tan(90) and tan(270)? That’s a good question, what on
earth are those equal to?
When our angle 𝑥 is at 90 degrees or 270 degrees, our adjacent length 𝑎 = 0.
Thus,
tan(90) =
𝑜𝑝
𝑎
=
𝑜𝑝
0
= ? ? ? (𝑀𝑎𝑡ℎ 𝑒𝑟𝑟𝑜𝑟) =
−𝑜𝑝
0
= tan(270).
In other words, there are actually no output values for the tangent function for the inputs
𝑥 = 90 and 𝑥 = 270.
This will become clearer when we draw the graph of the tangent function in the next section.
We have now defined the sine, cosine and tangent functions for the real numbers 𝑥 such that
0 ≤ 𝑥 ≤ 360. We will now define them on all real numbers ≥ 0.
Did you notice how sin(360) = sin(0) , cos(360) = cos(0) and tan(360) = tan(0)?
This is because once we have gone around 360 degrees, we have travelled in a circle, and we
are back where we started.
Hence, everything just repeats itself:
Hence, for 0 ≤ 𝑥 ≤ 360, we have:
sin(𝑥) = sin(360 + 𝑥) = sin(2 × 360 + 𝑥) = sin(3 × 360 + 𝑥) = ⋯
cos(𝑥) = cos(360 + 𝑥) = cos(2 × 360 + 𝑥) = cos(3 × 360 + 𝑥) = ⋯
tan(𝑥) = tan(360 + 𝑥) = tan(2 × 360 + 𝑥) = tan(3 × 360 + 𝑥) = ⋯
This defines our three functions for all real numbers 𝑥 ≥ 0.
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Lastly, we define our three trigonometric functions for negative real number inputs, 𝑥 < 0.
To do so, let us put a meaning towards a “negative angle.” After-all, the real numbers that we
are putting into our trigonometric functions are angles of triangles.
For 0 < 𝑥 < 90 it is evident from the above diagram that the following hold:
sin(−𝑥) =
−𝑜𝑝
𝑜𝑝
= −
= −sin(𝑥)
ℎ
ℎ
cos(−𝑥) =
tan(−𝑥) =
𝑎
= 𝑐𝑜𝑠(𝑥)
ℎ
−𝑜𝑝
𝑜𝑝
= −
= −tan(𝑥)
𝑎
𝑎
And, in fact, I’ll leave it as an exercise to show that for all 𝑥 ≥ 0 the following hold:
sin(−𝑥) = −sin(𝑥)
cos(−𝑥) = 𝑐𝑜𝑠(𝑥)
tan(−𝑥) = −tan(𝑥)
(Of course the values of 𝑥 where tangent is undefined remain undefined.)
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Exercises
1. Find the unknown lengths in the following triangles:
2. Show that tan(45) = 1.
Hint: A right-angled isosceles triangle has two equal angles.
3. Show that sin(45) =
1
.
√2
Hint: Again consider a right-angled, isosceles triangle and use Pythagoras theorem.
4. Show that the cosine function is properly defined on 0 < 𝑥 < 90.
5. Show that cos(45) =
1
.
√2
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6. Show that for 270 < 𝑥 < 360 we have:
sin(𝑥) =
−𝑜𝑝
𝑜𝑝
= −
= −sin(360 − 𝑥)
ℎ
ℎ
cos(𝑥) =
tan(𝑥) =
𝑎
= cos(360 − 𝑥)
ℎ
−𝑜𝑝
𝑜𝑝
= −
= −tan(360 − 𝑥)
𝑎
𝑎
7. For 𝑥 = 90, 180, 270 and 360, deduce the values of sin(𝑥) , cos(𝑥) and tan(𝑥).
8. For the following triangles, find the values of the unknown sides as functions in terms of
the angle 𝑥:
9. Show that for all real numbers 𝒙 (except isolated values for tangent) the following hold:
sin(𝑥) = sin(180 − 𝑥), cos(𝑥) = −cos(180 − 𝑥), tan(𝑥) = tan(𝑥 − 180).
10. Show that 𝑠𝑖𝑛2 (𝑥) + 𝑐𝑜𝑠 2 (𝑥) = 1
(Here, for example, 𝑠𝑖𝑛2 (𝑥) = (sin(𝑥))2 )
Hint: Draw a triangle with hypotenuse 1 and use Pythagoras’ Theorem.