Download Inverse Trigonometric Functions

Inverse Trigonometric Functions
Recall from Chapter 1 that some functions f (x ) have inverse functions (written f −1 ( x) and read as
‘f-inverse’). The two functions are symmetrical to one another across the 45-degree line, and they
have the effect of ‘undoing’ the action of one another. Their points are the same except their x and
y values are exchanged.
A minimum requirement for the existence of an inverse function for f (x ) is that f (x ) itself must
pass the horizontal line test (HLT). This is required because when we turn f (x ) on its side (i.e.
reflect across the 45-degreeline), we want the inverse graph to also be a function, too (passing the
VLT). Such a function is called ‘one-to-one’. Functions such as parabolas are not one-to-one, so to
remedy this problem we take just a portion of a parabola so that it is one-to-one and then we can
find its inverse function.
Given the function f ( x) = sin x , we graph it and see that it is not one-to-one. If we want to create
an inverse function for sin x, we need to take just a portion of the sine graph so that this portion is
one-to-one. To do this we take the portion over the interval − π2 ≤ x ≤ π2 , corresponding to
quadrants 1 and 4:
We can now define the inverse function for f ( x) = sin x . It is called ‘sine-inverse’ and written
f −1 ( x) = sin −1 x . Do not confuse the exponent “-1” for reciprocals. In this context it always means
the inverse function. Sometimes this function is also called the arcsine function.
Notice how the x and y-values switch on the axes. The sine-inverse function works this way: since
sin( π4 ) = 22 , then sin −1 ( 22 ) = π4 , for example. The outputs of the inverse-trig functions are
angles (in radians by default).
The inverse-cosine and inverse-tangent functions are also defined similarly. Since the cos x and tan
x functions are not one-to-one either, we have to take just a portion of their graphs to define their
inverses. For the cos x graph, we take just the portion from 0 ≤ x ≤ π (do you see why?). These
correspond to quadrants 1 and 2. For the inverse-tangent graph, we take just the portion on
− π2 ≤ x ≤ π2 , which correspond to quadrants 1 and 4. See the graphs in your text for samples.
It’s important to know that the inverse-sine (or cosine, etc) function gives back just one value for an
input x. You need to use symmetry and a knowledge of quadrants to get any other possible values.
Example: In degree mode, we calculate y = sin −1 (0.75) and we get 48.59 degrees. In other words,
when a ray elevated at 48.59 degrees is drawn on the unit circle, its y-coordinate (the sin of 48.59) is
0.75. However, there is one other answer! There is another point on the unit circle whose ycoordinate is 0.75 – it lies in the second quadrant, so we use symmetry and reflections to determine
that the other answer must be 131.41 degrees.
The same process is used for the inverse-cosine graph, except that we refer everything based off the
x-axis, to generate the second result. For the inverse-tangent function, we just get the single value.
Find both solutions to each of these problems (use degrees):
cos −1 (0.92) , sin −1 (−0.34) , cos −1 (−0.15) , sin −1 (0.21)
Find both solutions (in exact radians) to cos −1 (−0.5) .
The inverse-tangent function will give back answers in the 1st and 4th quadrants. The other
solutions will exist in the third and second quadrants, respectively. Just extend the ray all the way
through and use symmetry.
For example: The expression tan −1 (2) says “what ray gives a slope of 2?”. In degrees, we get
63.43 degrees (use your calculator). This is in quadrant 1. If you extend this ray into quadrant 3,
you can add 180 degrees to your answer, so another possible result to this question is 243.43
degrees.
Find both results to these questions (in degrees): tan −1 (0.5) , tan −1 (−1.2) , tan −1 (−0.02) .
Find both results in radians to tan −1 (−
3
3
) . Hint: use your 1st-quadrant table knowledge.
Combining the inverse trigonometric functions with the trigonometric functions.
The inverse-trigonometric functions are meant to ‘cancel’ out their respective trigonometric
function:
sin −1 (sin( x )) = x, − π2 ≤ x ≤
π
2
(Q4 and Q1)
sin(sin −1 ( x)) = x, − 1 ≤ x ≤ 1
cos −1 (cos( x)) = x, 0 ≤ x ≤ π (Q1 and Q2)
cos(cos −1 ( x)) = x, − 1 ≤ x ≤ 1
tan −1 (tan( x )) = x, − π2 ≤ x ≤ π2 (Q4 and Q1)
tan(tan −1 ( x)) = x, all x
The problem, of course, is remembering when these cancellations are defined, and whether the
inverse part comes first or second. It can be kind of a hassle.
Case I: The inverse-trig function is performed last (i.e. is the ‘outside’ of the two functions).
In this case, just remember which quadrants the inverse-trig function is defined for. For example,
sin −1 (sin( π4 )) = π4 since π4 is in quadrant 1. However, sin −1 (sin( 34π )) = π4 since 34π is in Q2. In this
case, the inverse-sine gives the equivalent Q1 answer.
Try these (remember which quadrant the angle is in):
sin −1 (sin( π6 )) , sin −1 (sin( − π3 )) , cos −1 (cos( π7 )) , cos −1 (cos( − 25π )) , tan −1 (tan( 10π )) , tan −1 (tan( 116π ))
Case II: The trig function is performed last.
In this case, the cancellation is true, as long as the inverse-trig function is defined on the inside. For
example, sin(sin −1 (0.7)) = 0.7 since 0.7 is between -1 and 1, but sin(sin −1 (1.7)) = ∅ , since
sin −1 (1.7) is not defined. The equation tan(tan −1 ( x)) = x is always true.
Try these:
sin(sin −1 (0.25)) , cos(cos −1 ( 109 )) , sin(sin −1 (2)) , cos(cos −1 (−1.1)) , tan(tan −1 (3.456)) .
Case III: You are combining different inverses with trig functions.
An example would be cos(sin −1 ( 23 )) . In this case, we know that “ sin −1 ( 23 ) ” is some angle whose
sine is 2/3. We set of a triangle, labeling the opposite 2 and the hypotenuse 3. By Pythagoras’
theorem, the adjacent side is 5 . Since cosine of this angle is adjacent over hypotenuse, we have
cos(sin −1 ( 23 )) =
5
3
.
If the inverse-trig function is on the outside, you want to try to get the inner function to be the same.
For example, if we had sin −1 (cos(70 o )) , we’d want to get that interior cosine into its equivalent
sine. We use the fact that sin( x ) = cos(90 − x ) (and vice-versa), so cos( 70) = sin( 20) . Therefore,
sin −1 (cos(70 o )) = sin −1 (sin 20) = 20 degrees.
Try these:
sin −1 (cos 30) , cos −1 (sin 15) .