INVERSE TRIGONOMETRIC FUNCTIONS

The range of y = arcsin x

The range of y = arctan x

The range of y = arccos x

The range of y = arcsec x

The inverse relations

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 13.)

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 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 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.

Thus,

arcsin ½ =

π
6

if and only if ½ = sin

π
6

Corresponding to each trigonometric function, there is its inverse function.

arcsin x,

arccos x,

arctan x,

arccsc x,

arcsec x,.

arccot x.

In each one, we are given the value x of the trigonometric function. We are to name the radian angle that has that value.

Problem 1.

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Do the problem yourself first!

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.

d) arccot 1 =

π
4

if and only if

Example 1. Evaluate arcsin

-- "the angle whose sine is

Solution.

arcsin

π
4

because

= sin

π
4

(Topic 4.)

The range of y = arcsin x

But

π
4

is not the only angle whose sine is

is the sine of every first

and second quadrant angle whose corresponding acute angle is

π
4

sin

3π
4

sin (

π
4

+ 2π

) =

And so on.

For the function y = arcsin x to be single-valued, we must restrict the values of the angle y. How will we do that? We will restrict them to those angles that have the smallest absolute value.

π
4

is the angle of smallest absolute value whose sine is

. It is a first

quadrant angle betwenn 0 and

π
2

Example 2. Evaluate arcsin (−

Solution. Angles whose sines are negative fall in the 3rd and 4th quadrants. The angle of smallest absolute value falls in the 4th quadrant

between 0 and −

π
2

It is −

π
4

arcsin (−

) = −

π
4

The range, then, of the function y = arcsin x will be angles that fall in

the 1st and 4th quadrants, between −

π
2

and

π
2

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,

arcsin (−x)

−arcsin x.

The angle whose sine is −x is simply the negative of the angle whose sine is x.

To see that, look here:

	=	θ.

	=	−θ.

That is,

arcsin(−x)	=	−arcsin x.

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.)

(Topic 5.)

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 15.)

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.

	=	−θ.

	=	θ.

Therefore,

arctan(−x)	=	−arctan x.

Problem 3. Evaluate the following.

a) arctan 1 =

b) arctan (−1) =

c) tan⁻¹

d) tan⁻¹(−

)

e) arctan 0 =

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

Solution. The radian angle whose cosine is ½ is

π
3

(60°).

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
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 =

b) arccos (−1) =

c) cos⁻¹

d) cos⁻¹(−

) =

e) arccos 0 =

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

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.

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:

arcsin (x − 1) =