The Mathematics 11
Competency Test
Inverse Trigonometric Functions
In the preceding document in these notes, we defined
and illustrated the use of the principal trigonometric
ratios
sin A =
a
c
sin B =
b
c
cos A =
b
c
cos B =
a
c
a
b
tan B =
tan A =
B
c
A
a
C
b
b
a
for a right triangle.
The reason sin A, cos A and tan A are called trigonometric ratios is because their values are
values of ratios of lengths of sides of right triangles. However, as pointed out earlier, the actual
values of sin A, cos A and tan A are also determined by the value of angle A itself.
Mathematicians say that sin A is a function of A, and so it is common to refer to sin A, cos A, and
tan A, as trigonometric functions of the angle A. You could think of the “sine function” as a
thing that takes the value of an angle in a right triangle, and gives you back the value of the ratio
of the lengths of two sides of that triangle. The sine function is like a ratio producing machine –
you feed it the value of an angle and it spits out the value of the corresponding sine ratio. The
same is true, of course, for the other trigonometric functions. Each spits out the value of the
corresponding trigonometric ratio when you feed it the value of a specific angle in a right triangle.
However, when we solve right triangles, we often have enough information to calculate the ratio
of the length of two sides, but we need to determine the value of one of the acute angles in the
triangle. For this, we need functions which do the opposite of sin A, cos A, and tan A – that is,
they must start with the value of the ratio of the lengths of two sides of a right triangle and
produce one of the acute angles. We need a set of functions which produce the actual value of
angle A from known values of sin A, cos A, or tan A, respectively. Such functions exist and are
programmed into scientific calculators. They are called inverse trigonometric functions, and
are defined by
sin-1(x) = the angle whose sine ratio equals x
cos-1(x) = the angle whose cosine ratio equals x
and
tan-1(x) = the angle whose tangent ratio equals x.
Some older books and calculators use the names arcsin, arccos, and arctan in place of sin-1,
cos-1, and tan-1, respectively. Note that the superscripts ‘-1’ are labels and not powers or
exponents. In particular, it is definitely NOT the case that these functions are simply the
reciprocals of the trigonometric functions:
sin−1( x ) is not the same thing as
David W. Sabo (2003)
1
sin x
Inverse Trigonometric Functions
Page 1 of 4
and
cos−1( x ) is not the same thing as
1
cos x
tan−1( x ) is not the same thing as
1
tan x
and
In fact, in the above three statements, the references to sin x, cos x, and tan x are nonsense,
because x stands for the value of a ratio of the lengths of two sides of a right triangle, whereas
the information required by the sin, cos and tan functions are angles. Since x is not an angle in
this context, sin x is meaningless – it would be a little bit like trying to calculate the sine of an
elephant, for instance. So, sin-1x takes the value, x, of a sine ratio and produces the
corresponding angle, in contrast to sin A, which takes the value, A, of an angle and produces the
corresponding value of the sine ratio. In very loose terms (though true if we are speaking just
about right triangles), we can represent the relationship between the two sets of functions as:
sin(sin-1x) = x
sin-1(sin A) = A
cos(cos-1x) = x
cos-1(cos A) = A
tan(tan-1x) = x
tan-1(tan A) = A
Here, for instance, sin(sin-1x) means “calculate the sin for the angle sin-1x”, and sin-1(sin A) means
“calculate the inverse sine for the value of sin A.” These “identities” are getting pretty deep into
the subject of functional notation and inverse functions, and you needn’t puzzle over them too
long. But you should be able to see that when the mathematical notation is replaced by the
English equivalents of the type given in quotation marks in the preceding sentence, these
relations make sense.
(All the same, this notation is not as complicated as it might look to you at first, so maybe it’s worth pausing
here to puzzle one of these out. The quantity
-1
sin(sin x)
can be expanded to
sin(the angle whose sine ratio equals x).
So, what is the sine of the angle whose sine equals x? It has to be x! That is
-1
sin(sin x) = x
as stated in the first line above. If you still can’t quite see this, trying a slightly different question: What is
the first name of a person whose first name is John?
Similarly, the quantity
-1
sin (sin A)
can be expanded to
“What is the angle whose sine is equal to the sine of angle A?”
The answer here is obviously A, giving
-1
sin (sin A) = A
David W. Sabo (2003)
Inverse Trigonometric Functions
Page 2 of 4
as stated on the right in the first line above.
This may seem like a lot of time spent on a rather obscure point, but if you’ve come to understand the point
here, you’ve learned a lot about the way to understand mathematical notation.)
Now, back to the main discussion of inverse trigonometric functions. Two very important points
must be kept in mind when working with or using these functions.The first important point of this
discussion is that sin-1, cos-1 and tan-1 are not related to sine, cosine and tangent in a simple
algebraic manner, but are a quite distinct set of functions. On scientific calculators, they are
usually accessed by first pressing the 2nd or INV key and then the sin, cos or tan key.
The second important point is that when you use the sin, cos, or tan key on your calculator, the
calculator is expecting you to enter the value
of an angle, and it will produce the value of the
corresponding ratio of two sides of the right
triangle. If you use the sin-1, cos-1 or tan-1
keys on your calculator, it is expecting you to
13
enter the value of the appropriate ratio of the
5
lengths of two sides of the right triangle, and it
will produce the corresponding angle.
θ
12
Example 1: Determine the value of the angle
θ in the right triangle shown in the sketch.
Round your answer in degrees to two decimal places.
solution:
This is a right triangle, so the definitions of sin θ, cos θ, and tan θ that we’ve been using apply
here. For this triangle
sinθ =
5
12
, cosθ = , and
13
13
tanθ =
5
12
So, formally, we can write:
since
sinθ =
5
,
13
then
 5 

 13 
θ = sin−1 
To get an actual numerical value for θ, first make sure your calculator is in “degree” mode, and
then invoke the sin-1 function for the value “5 ÷ 13.” (The specific keystrokes will depend on the
type of calculator you have. On some, pressing 2nd, sin, (, 5, ÷, 13, ), Enter, will do the job. On
others, you may have to first calculate the “5 ÷ 13” and then invoke the inverse sine function.)
On my calculator, I get
 5 
≅ 22.619864950

 13 
θ = sin−1 
Rounding to two decimal places gives the final answer
θ = 22.620
David W. Sabo (2003)
Inverse Trigonometric Functions
Page 3 of 4
You should check to make sure that you get exactly the same value using
 12 

 13 
θ = cos−1 
and
 5 
.
 12 
θ = tan−1 
Example 2: Determine the angles A
and B for the triangle shown in the
figure to the right. Verify that your
answers add up to 900, as they must do
for a right triangle.
solution:
Since we are given the lengths of the
legs of this right triangle, it is the tangent
ratios that we must use:
56
90
A
B
tan A =
90
56
⇒
 90 
A = tan−1   ≅ 58.10920
 56 
tan B =
56
90
⇒
 56 
B = tan−1   ≅ 31.89080
 90 
and
Now, it is easy to confirm that
A + B ≅ 58.10920 + 31.89080 = 90.00000.
Remarks:
• the description of inverse trigonometric functions given above is adequate for right triangle
problems. Some extra details must be taken into account when you deal with angles outside of
the 00 and 900 range. But this situation does not arise in the BCIT Mathematics 11 Competency
Test.
• the way that the sine and cosine ratios are defined means that for right triangles, sines and
cosines have values that range from 0 to 1 in magnitude. If you try to compute sin-1(x) or cos-1(x)
for values of x outside of this range, your calculator will give an error. What your request
amounts to in such a situation is asking for an angle in an impossible right triangle (one in which
the hypotenuse is shorter than one of the other sides of the triangle).
David W. Sabo (2003)
Inverse Trigonometric Functions
Page 4 of 4