21
INVERSE TRIGONOMETRIC FUNCTIONS
THE ANGLES in theoretical work will be in radian measure. Thus if
| we are given a radian angle, | π 6 |
for example, then we can evaluate a |
function of it.
| sin | π 6 |
= | ½. |
(Topic 15.)
Inversely, if we are given a value of the function -- ½ -- then the challenge is to name the angle.
sin x = ½.
"The sine of what angle is equal to ½?"
We could answer:
| "The angle whose sine is ½ is | π 6 |
." |
The algebraic abbreviation for that sentence is
| "arcsin ½ = | π 6 |
." |
arcsin x is called the inverse sine function. It is the angle whose sine is x.
Strictly, arsin x is the arc whose sine is x. Because in the unit circle, the length of that arc is the radian measure. Topic 16.
Thus the inverse of the function
y = sin x
"The number y is the sine of the radian angle x"
is
y = arcsin x.
"y is the radian angle whose sine is the number x."
Corresponding to each trigonometric function, there is its inverse function:
arcsin x,
arccos x,
arctan x,
arccsc x,
arcsec x,.
arccot x.
The inverse relations
If we put
f(x) = sin x
and
g(x) = arcsin x,
then according to the definition of inverse functions (Topic 19 of Precalculus):
f(g(x)) = x and g(f(x)) = x.
sin(arcsin x) = x and arcsin(sin x) = x.
In particular,
| arcsin x | = | y | |
| implies, on taking the inverse function -- the sine -- of both sides: | |||
| x | = | sin y. | |
Problem 1. What does each of the following imply?
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Do the problem yourself first!
a) arccos N = θ. N = cos θ.
b) arctan t = β. t = tan β.
c) arcsec u = a. u = sec a.
| d) arccot 1 = | π 4 |
. | 1 | = cot | π 4 |
. |
| Example 1. Evaluate arcsin |  2 | -- "the angle whose sine is | 2 |
." |
| Solution. | 2 |
is the sine of what angle? | |
2 |
= | sin | π 4 |
(Topic 6). | |
That is,
| arcsin | 2 |
= | π 4 |
. | ||
The range of y = arcsin x
| But | π 4 |
is not the only angle whose sine is | 2 |
. | 2 |
is the sine of every 1st and | ||
| 2nd quadrant angle whose corresponding acute angle is | π 4 |
 |
||
| sin | 3π 4 |
= | 2 |
. |
| sin ( | π 4 |
+ 2π | ) = | 2 |
. |
And so on.
For the function y = arcsin x to be single-valued, then, we must restrict the values of y. How will we do that? We will restrict them to those angles that have the smallest absolute value.
In that same way, we will restrict the range of each inverse trigonometric function. (Topic 3 of Precalculus.)
| π 4 |
is the angle of smallest absolute value whose sine is | 2 |
. |
| Example 2. Evaluate arcsin (− | 2 |
). |
Solution. Angles whose sines are negative fall in the 3rd and 4th quadrants. (Topic 17.) The angle of smallest absolute value is in the 4th
| quadrant. It is the angle − | π 4 |
. | ||
| arcsin (− | 2 |
) = − | π 4 |
. | ||
For angles whose sine is negative, we always choose a 4th quadrant angle. In fact,
| arcsin (−x) | = | −arcsin x. | ||
| arcsin(−½) | = | −arcsin ½ | ||
| = | − | π 6 |
. | |
*
To see that arcsin(−x) = −arcsin x, look here:
 |
= | θ. |
 |
= | −θ. |
| That is, | ||
| arcsin(−x) | = | −arcsin x. |
Here, then, is the range of the function y = arcsin x.
To restrict the range of arcsin x is equivalent to restricting the domain of sin x to those same values. This will be the case with all the restricted ranges that follow.
Another notation for arcsin x is sin−1x. Read: "The inverse sine of x." −1 here is not an exponent (See Topic 19 of Precalculus.)
Problem 2. Evaluate the following in radians.
a) arcsin 0 = 0. (Topic 17.)
b) arcsin 1 = π/2. (Topic 17.)
c) arcsin (−1) = −π/2. (Topic 17.)
 |
π/3. (Topic 7.) |
 |
−π/3. |
 |
−π/6. |
The range of y = arctan x
Similarly, we must restrict the range of y = arctan x. Like y = arcsin x, y = arctan x has its smallest absolute values in the 1st and 4th quadrants.
| Note that y -- the angle whose tangent is x -- must be greater than − | π 2 |
| and less than | π 2 |
. For, at those quadrantal angles, the tangent does not exist. |
(Topic 17.)
For angles whose tangent is postive, we choose a 1st quadrant angle. For angles whose tangent is negative, we choose a 4th quadrant angle. Like arcsin (−x),
arctan (−x) = −arctan x.
 |
= | −θ. |
 |
= | θ. |
| Therefore, | ||
| arctan(−x) | = | −arctan x. |
Problem 3. Evaluate the following.
| a) arctan 1 = | π 4 |
b) arctan (−1) = | − | π 4 |
c) tan−1 |
= | π 3 |
d) tan−1(−) |
= | − | π 3 |
| e) arctan 0 = | 0 | f) |  |
= | − | π 6 |
The range of y = arccos x
The values of y = arccos x will have their smallest absolute values when y falls within the 1st or 2nd quadrants.
Example 3. Evaluate
a) arccos ½
| Solution. The radian angle whose cosine is ½ is | π 3 |
(60°). |
b) arccos (−½)
Solution. An angle x whose cosine is negative falls in the 2nd quadrant.
And the cosine of a 2nd quadrant angle is the negative of the cosine of its supplement. (Topic 18.) This implies:
An angle θ whose cosine is −x is the supplement
of the angle whose cosine is x.
arccos (−x) = π − arccos x.
Therefore,
| arccos (−½) | = | π − arccos ½ |
| = |  |
|
| = | 2π 3 |
Problem 4. Evaluate the following.
| a) arccos 1 = | 0 | b) arccos (−1) = | π |
| c) cos−1 | 2 |
= | π 4 |
d) cos−1(− | 2 |
) = | 3π 4 |
| e) arccos 0 = | π 2 |
f) |  |
= | 5π 6 |
The range of y = arcsec x
In calculus, sin−1x, tan−1x, and cos−1x are the most important inverse trigonometric functions. Nevertheless, here are the ranges that make the rest single-valued.
Similarly for y = arccsc x.
Next Topic: Trigonometric identities
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