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Evaluating compositions of the form f ( f −1 ( y )) and f −1 ( f ( x ))

For any trigonometric function, f ( f 1 ( y ) ) = y for all y in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of f was defined to be identical to the domain of f 1 . However, we have to be a little more careful with expressions of the form f 1 ( f ( x ) ) .

Compositions of a trigonometric function and its inverse

sin ( sin 1 x ) = x for 1 x 1 cos ( cos 1 x ) = x for 1 x 1 tan ( tan 1 x ) = x for < x <


sin 1 ( sin x ) = x only for  π 2 x π 2 cos 1 ( cos x ) = x only for  0 x π tan 1 ( tan x ) = x only for  π 2 < x < π 2

Is it correct that sin 1 ( sin x ) = x ?

No. This equation is correct if x belongs to the restricted domain [ π 2 , π 2 ] , but sine is defined for all real input values, and for x outside the restricted interval, the equation is not correct because its inverse always returns a value in [ π 2 , π 2 ] . The situation is similar for cosine and tangent and their inverses. For example, sin 1 ( sin ( 3 π 4 ) ) = π 4 .

Given an expression of the form f −1 (f(θ)) where f ( θ ) = sin θ ,   cos θ ,  or  tan θ , evaluate.

  1. If θ is in the restricted domain of f ,  then  f 1 ( f ( θ ) ) = θ .
  2. If not, then find an angle ϕ within the restricted domain of f such that f ( ϕ ) = f ( θ ) . Then f 1 ( f ( θ ) ) = ϕ .

Using inverse trigonometric functions

Evaluate the following:

  1. sin 1 ( sin ( π 3 ) )
  2. sin 1 ( sin ( 2 π 3 ) )
  3. cos 1 ( cos ( 2 π 3 ) )
  4. cos 1 ( cos ( π 3 ) )
  1. π 3  is in  [ π 2 , π 2 ] , so sin 1 ( sin ( π 3 ) ) = π 3 .
  2. 2 π 3  is not in  [ π 2 , π 2 ] , but sin ( 2 π 3 ) = sin ( π 3 ) , so sin 1 ( sin ( 2 π 3 ) ) = π 3 .
  3. 2 π 3  is in  [ 0 , π ] , so cos 1 ( cos ( 2 π 3 ) ) = 2 π 3 .
  4. π 3  is not in  [ 0 , π ] , but cos ( π 3 ) = cos ( π 3 ) because cosine is an even function.
  5. π 3  is in  [ 0 , π ] , so cos 1 ( cos ( π 3 ) ) = π 3 .
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Evaluate tan 1 ( tan ( π 8 ) ) and tan 1 ( tan ( 11 π 9 ) ) .

π 8 ; 2 π 9

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Evaluating compositions of the form f −1 ( g ( x ))

Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form f 1 ( g ( x ) ) . For special values of x , we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is θ , making the other π 2 θ . Consider the sine and cosine of each angle of the right triangle in [link] .

An illustration of a right triangle with angles theta and pi/2 - theta. Opposite the angle theta and adjacent the angle pi/2-theta is the side a. Adjacent the angle theta and opposite the angle pi/2 - theta is the side b. The hypoteneuse is labeled c.
Right triangle illustrating the cofunction relationships

Because cos θ = b c = sin ( π 2 θ ) , we have sin 1 ( cos θ ) = π 2 θ if 0 θ π . If θ is not in this domain, then we need to find another angle that has the same cosine as θ and does belong to the restricted domain; we then subtract this angle from π 2 . Similarly, sin θ = a c = cos ( π 2 θ ) , so cos 1 ( sin θ ) = π 2 θ if π 2 θ π 2 . These are just the function-cofunction relationships presented in another way.

Given functions of the form sin 1 ( cos x ) and cos 1 ( sin x ) , evaluate them.

  1. If x  is in  [ 0 , π ] , then sin 1 ( cos x ) = π 2 x .
  2. If x  is not in  [ 0 , π ] , then find another angle y  in  [ 0 , π ] such that cos y = cos x .
    sin 1 ( cos x ) = π 2 y
  3. If x  is in  [ π 2 , π 2 ] , then cos 1 ( sin x ) = π 2 x .
  4. If x  is not in [ π 2 , π 2 ] , then find another angle y  in  [ π 2 , π 2 ] such that sin y = sin x .
    cos 1 ( sin x ) = π 2 y

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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