For the function y = arcsin x to be single-valued, 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.

π
4

is the angle of smallest absolute value whose sine is

Example 2. Evaluate arcsin (−

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

It is −

π
4

arcsin (−

) = −

π
4

For an angle whose sine is negative, we must choose a 4th quadrant angle. In 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.

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 15.)

b) arcsin 1 = π/2. (Topic 15.)

c) arcsin (−1) = −π/2. (Topic 15.)

(Topic 5.)

The range of y = arctan x

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.

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.

Similarly for y = arccsc 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.

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) =