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Evaluating the composition of an inverse sine with a cosine

Evaluate sin 1 ( cos ( 13 π 6 ) )

  1. by direct evaluation.
  2. by the method described previously.
  1. Here, we can directly evaluate the inside of the composition.
    cos ( 13 π 6 ) = cos ( π 6 + 2 π )                 = cos ( π 6 )                 = 3 2

    Now, we can evaluate the inverse function as we did earlier.

    sin 1 ( 3 2 ) = π 3
  2. We have x = 13 π 6 , y = π 6 , and
    sin 1 ( cos ( 13 π 6 ) ) = π 2 π 6 = π 3        
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Evaluate cos 1 ( sin ( 11 π 4 ) ) .

3 π 4

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

To evaluate compositions of the form f ( g 1 ( x ) ) , where f and g are any two of the functions sine, cosine, or tangent and x is any input in the domain of g 1 , we have exact formulas, such as sin ( cos 1 x ) = 1 x 2 . When we need to use them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagoras’s relation between the lengths of the sides. We can use the Pythagorean identity, sin 2 x + cos 2 x = 1 , to solve for one when given the other. We can also use the inverse trigonometric functions to find compositions involving algebraic expressions.

Evaluating the composition of a sine with an inverse cosine

Find an exact value for sin ( cos 1 ( 4 5 ) ) .

Beginning with the inside, we can say there is some angle such that θ = cos 1 ( 4 5 ) , which means cos θ = 4 5 , and we are looking for sin θ . We can use the Pythagorean identity to do this.

sin 2 θ + cos 2 θ = 1 Use our known value for cosine . sin 2 θ + ( 4 5 ) 2 = 1 Solve for sine . sin 2 θ = 1 16 25 sin θ = ± 9 25 = ± 3 5

Since θ = cos 1 ( 4 5 ) is in quadrant I, sin θ must be positive, so the solution is 3 5 . See [link] .

An illustration of a right triangle with an angle theta. Oppostie the angle theta is a side with length 3. Adjacent the angle theta is a side with length 4. The hypoteneuse has angle of length 5.
Right triangle illustrating that if cos θ = 4 5 , then sin θ = 3 5

We know that the inverse cosine always gives an angle on the interval [ 0 , π ] , so we know that the sine of that angle must be positive; therefore sin ( cos 1 ( 4 5 ) ) = sin θ = 3 5 .

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Evaluate cos ( tan 1 ( 5 12 ) ) .

12 13

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Evaluating the composition of a sine with an inverse tangent

Find an exact value for sin ( tan 1 ( 7 4 ) ) .

While we could use a similar technique as in [link] , we will demonstrate a different technique here. From the inside, we know there is an angle such that tan θ = 7 4 . We can envision this as the opposite and adjacent sides on a right triangle, as shown in [link] .

An illustration of a right triangle with angle theta. Adjacent the angle theta is a side with length 4. Opposite the angle theta is a side with length 7.
A right triangle with two sides known

Using the Pythagorean Theorem, we can find the hypotenuse of this triangle.

       4 2 + 7 2 = hypotenuse 2 hypotenuse = 65

Now, we can evaluate the sine of the angle as the opposite side divided by the hypotenuse.

sin θ = 7 65

This gives us our desired composition.

sin ( tan 1 ( 7 4 ) ) = sin θ                        = 7 65                        = 7 65 65
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Evaluate cos ( sin 1 ( 7 9 ) ) .

4 2 9

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Finding the cosine of the inverse sine of an algebraic expression

Find a simplified expression for cos ( sin 1 ( x 3 ) ) for 3 x 3.

We know there is an angle θ such that sin θ = x 3 .

sin 2 θ + cos 2 θ = 1 Use the Pythagorean Theorem .   ( x 3 ) 2 + cos 2 θ = 1 Solve for cosine .              cos 2 θ = 1 x 2 9                  cos θ = ± 9 x 2 9 = ± 9 x 2 3

Because we know that the inverse sine must give an angle on the interval [ π 2 , π 2 ] , we can deduce that the cosine of that angle must be positive.

cos ( sin 1 ( x 3 ) ) = 9 x 2 3
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Practice Key Terms 6

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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