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# 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

sin−1x. The inverse sine

The inverse relations

THE ANGLES in theoretical work will be in radian measure. Thus if we are given a radian angle,  for example, then we can evaluate a function of it.

 sin π6 = ½.

(Topic 13.)

Inversely, if we are given that the value of the sine function is ½, then the challenge is to name the radian angle x.

sin x = ½.

"The sine of what angle is equal to ½?"

 "The angle whose sine is ½  is π6 ."

The algebraic abbreviation for that sentence is:

 "arcsin ½ = π6 ."

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

is

y = arcsin x.

Their inverse relation is as follows:

arcsin x = θ  if and only if  x = sin θ.

Thus,

 arcsin ½ = π6 if and only if   ½ = sin π6 .

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.

Do the problem yourself first!

a)  arctan t = β  if and only if  ttan 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  = π4 if and only if 1 = cot π4 .

Example 1.   Evaluate arcsin -- "the angle whose sine is ."

 Solution. arcsin 2 = π4 because 2 = sin π4 . (Topic 4.)

The range of y = arcsin x

But is not the only angle whose sine is .   is the sine of every first and second quadrant angle whose corresponding acute angle is .

 sin 3π 4 = 2 .
 sin ( π4 + 2π ) = 2 .

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.

They are called the principal values of arcsin x.

is the angle of smallest absolute value whose sine is .  It is the first quadrant angle between 0 and .

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 −.

 It is  − π4 .
 arcsin (− 2 ) = − π4 .

The range, then, of the function  y = arcsin x will be angles that fall in the 1st and 4th quadrants, between − and .

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 quadrant.  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.

sin−1x. 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.)

 π/3.  (Topic4.)
 −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 − and less than . 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, then, as with arcsin (−x), we choose a 4th quadrant angle:

arctan (−x) = −arctan x.

 = −θ. = θ. Therefore, arctan(−x) = −arctan x.

Problem 3.   Evaluate the following.

 a)  arctan 1 = π4 b)  arctan (−1) = − π4
 c)  tan−1 = π3 d)  tan−1(−) = − π3
 e)  arctan 0 = 0 f) = − π6

The range of y = arccos x

Example 3.   Evaluate arccos ½.

 Solution. arccos ½ = π3 .
 The radian angle whose cosine is ½ is π3 (60°).

Problem 4.   Why is this not true?

arccos (−½) = −.

is a 4th quadrant angle. And in the 4th quadrant, the cosine is positive.

An angle whose cosine is negative will fall in the 2nd quadrant, where it will have its smallest absolute value. (Topic 15.)

The cosine of a 2nd quadrant angle is the negative of the cosine of its corresponding acute angle, which is its supplement.

In other words:

The angle θ whose cosine is −x  is the supplement
of the angle whose cosine is x.

arccos (−x) = π − arccos x.

Example 4.   Evaluate arccos (−½).

Solution.   We have seen:

arccos ½ = .

Therefore, arccos (−½)  is the supplement of —which is the angle we must add to to equal π.

+ θ = π.

Now, is one-third of π. Therefore, its supplement will be two-

thirds of π:  .

θ =arccos (−½) = .

The range, then, of y = arccos x will be from 0 to π.

An angle whose cosine is positive will be a 1st quadrant angle; an angle whose cosine is negative will fall in the 2nd. It will be the supplement of the 1st quadrant angle.

Problem 5.   Evaluate the following.

 a)  arccos 1 = 0 b)  arccos (−1) = π
 c)  cos−1 2 = π4 d)  cos−1(− 2 ) = π − π4 = 3π 4
 e) = π6 f) = π − π6 = 5π 6
 g)  arccos 0 = π2

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, or 0. 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) = .

Solution.   By taking the sine of both sides, we can free the argument x − 1, and write immediately --

 x − 1 = sin = 2

Therefore,

 x = 1 + 2 .

Problem 6.   Solve for x:

tan (x + 2) = 1.

 x + 2 = arctan 1 = π4 .
 x = π4 − 2.

Problem 7.   Solve for x:

cos x2 = −1.

x2 = arccos −1 = π.

x = ±.

Theorem.   If

y = arcsec x,

then the product

sec y tan y  is never negative.

For, if  y = arcsec x,  then the angle y falls either in the first or second quadrants.  When angle y falls in the first quadrant, then both sec y and tan y are positive. Therefore their product is positive.

When angle y falls in the second quadrant, sec y and tan y are both negative, so that again their product is positive.

If y = 0, then tan y= 0, hence the product sec y tan y is 0.

Therefore, that product is never negative.

(This theorem is referenced in the proof of the derivative of y = arcsecx.)

Next Topic:  Trigonometric identities

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