INVERSE TRIGONOMETRIC FUNCTIONS
THE ANGLES in theoretical work will be in radian measure. Thus if
function of it.
Inversely, if we are given a value of the sine function -- ½ -- then the challenge is to name the radian angle x.
sin x = ½.
"The sine of what angle is equal to ½?"
We could answer:
The algebraic abbreviation for that sentence is
arcsin x is called the inverse sine function.
It is the angle whose sine is the number x.
Strictly, arcsin x is the arc whose sine is x. Because in the unit circle, the length of that arc is the radian measure. Topic 14.
The inverse of the function
y = sin x
y = arcsin x.
Their inverse relation is as follows:
arcsin x = θ if and only if x = sin θ.
Corresponding to each trigonometric function, there is its inverse function.
In each one, we are given the value x of the trigonometric function. We are to name the radian angle that has that value.
To see the answer, pass your mouse over the colored area.
a) arctan t = β if and only if t = tan β.
b) arcsec u = α if and only if u = sec α.
c) arccos 1 = 0 if and only if 1 = cos 0.
The range of y = arcsin x
And so on.
Solution. Angles whose sines are negative fall in the 3rd and 4th quadrants. The angle of smallest absolute value falls in the 4th quadrant
The range, then, of the function y = arcsin x will be angles that fall in
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.
Angles whose sines are positive will be 1st quadrant angles, while angle whose sines are negative will fall in the 4th. If fact,
The angle whose sine is −x is simply the negative of the angle whose sine is x.
To see that, look here:
The inverse sine
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) sin−1 0 = 0. (Topic 15.)
b) sin−1 1 = π/2. (Topic 15.)
c) sin−1 (−1) = −π/2. (Topic 15.)
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.
For an angle whose tangent is positive, we choose a 1st quadrant angle. For an angle whose tangent is negative, we choose a 4th quadrant angle. Like arcsin (−x),
arctan (−x) = −arctan x.
Problem 3. Evaluate the following.
The range of y = arccos x
The values of y = arccos x will have their smallest absolute values when y -- the angle -- falls in the 1st and 2nd quadrants. An angle whose cosine is positive will be a 1st quadrant angle; an angle whose cosine is negative will fall in the 2nd. (Topic 15.)
Example 3. Evaluate
a) arccos ½
b) arccos (−½)
Solution. An angle θ 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 16.) That implies:
An angle θ whose cosine is −x is the supplement
arccos (−x) = π − arccos x.
Problem 4. Evaluate the following.
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.
If x is positive, then the value of the inverse function is always a first quadrant angle. If x is negative, the value of the inverse will fall in the quadrant in which the direct function is negative. Thus if x is negative, arcsec x will fall in the 2nd quadrant, because that is where sec x is negative.
The only inverse function below in which x may be 0, is arccot x. arccot 0 = π/2.
Again, we restrict the values of y to those angles that have the smallest absolute value.
The inverse relations
If we put
f(x) = sin x
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.
By taking the inverse function of both sides, we have extracted, or freed, the argument x. (See Topic 19 of Precalculus, Extracting the argument.) That enables us to solve many trigonometric equations.
Example 4. Solve for x:
Solution. By taking the sine of both sides, we can free the argument x − 1, and write immediately --
Problem 5. Solve for x:
tan (x + 2) = 1.
Problem 6. Solve for x:
cos x² = −1.
x² = arccos −1 = π.
x = ±.
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Copyright © 2013 Lawrence Spector
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