This is an original lesson based on OpenStax Precalculus lesson 6.3.

Summary: In this section, you will:

- Evaluate compositions of inverse functions

Right triangles such as the one in **figure 1** can be used to simplify compositions of trigonometric functions such as sin(tan^{–1} *x*).

Previously we learned that in *f*(*x*) and *f*^{–1}(*x*) were inverses, then *f*(*f*^{–1}(*x*)) = *x* and *f*^{–1}(*f*(*x*)) = *x*. The same is true for trigonometric functions with an exception. The domain of the inverse functions must be applied.

If –1 ≤

*x*≤ 1 and \(–\frac{π}{2}\) ≤*y*≤ \(\frac{π}{2}\), then sin(sin^{–1}(*x*)) =*x*and sin^{–1}(sin(*y*)) =*y*If –1 ≤

*x*≤ 1 and 0 ≤*y*≤*π*, then cos(cos^{–1}(*x*)) =*x*and cos^{–1}(cos(*y*)) =*y*If

*x*is a real number and \(–\frac{π}{2}\) ≤*y*≤ \(\frac{π}{2}\), then tan(tan^{–1}(*x*)) =*x*and tan^{–1}(tan(*y*)) =*y*

Remember to be careful that the domain and range of the composition is maintained. Work through the composition from the inside out.

Evaluate a) \(\sin\left(\arcsin \frac{1}{2}\right)\), b) \(\cos\left(\cos^{–1} \frac{2π}{3}\right)\), c) tan(arctan –10).

- \(\sin\left(\arcsin \frac{1}{2}\right)\): arcsin is the inner function, and the domain of arcsin is –1 ≤
*x*≤ 1. \(\frac{1}{2}\) is in this domain. \(\arcsin\left(\frac{1}{2}\right) = \frac{π}{6}\), then find \(\sin\left(\frac{π}{6}\right) = \frac{1}{2}\). So \(\sin\left(\arcsin \frac{1}{2}\right) = \frac{π}{6}\). - \(\cos\left(\cos^{–1} \frac{2π}{3}\right)\): cos
^{–1}is the inner function, and the domain of cos^{–1}is –1 ≤*x*≤ 1. \( \frac{2π}{3} \approx 2\), so \(\frac{2π}{3}\) is outside of the domain and thus there is no solution for \(\cos\left(\cos^{–1} \frac{2π}{3}\right)\). - tan(arctan –10): arctan is the inner function, and the domain of arctan is any real number. –10 is a real number, so tan(arctan –10) = –10.

Evaluate a) sin(sin^{–1}(0.345)) and b) \(\cos\left(\cos^{–1} –\frac{2}{3}\right)\).

0.345; \(–\frac{2}{3}\)

Evaluate a) \(\arcsin\left(\sin \frac{π}{3}\right)\), b) \(\arccos\left(\cos \frac{5π}{4}\right)\), and c) tan^{–1}(tan *π*).

\(\arcsin\left(\sin \frac{π}{3}\right)\): Work from the inside out. \(\sin \frac{π}{3} = \frac{\sqrt{3}}{2}\) so

\(\arcsin\left(\sin \frac{π}{3}\right)\)

\(= \arcsin \frac{\sqrt{3}}{2}\)

\(= \frac{π}{3}\)\(\arccos\left(\cos \frac{5π}{4}\right)\): \(\cos \frac{5π}{4} = –\frac{\sqrt{2}}{2}\) so

\(\arccos\left(\cos \frac{5π}{4}\right)\)

\(= \arccos \left(–\frac{\sqrt{2}}{2}\right)\)

\(= \frac{3π}{4}\)(Remember the range of arccos is 0 ≤

*y*≤*π*.)tan

^{–1}(tan*π*): tan*π*= 0 sotan

^{–1}(tan*π*)

= tan^{–1}(0)

= 0(Remember the range of tan

^{–1}is \(–\frac{π}{2}\) ≤*y*≤ \(\frac{π}{2}\).)

Evaluate a) \(\arctan\left(\tan \frac{3π}{4}\right)\) and b) sin^{–1}(sin(–0.354)).

\(–\frac{π}{4}\); –0.354

Composition of trigonometric functions can also be solved using right triangles. Use the inner function to draw a right triangle, then use the triangle to evaluate the outer function.

- Draw a right triangle to represent the inner function. Two sides should be labeled.
- Use the Pythagorean Theorem to solve for the other side.
- Use the triangle to evaluate the outer trigonometric function.

Evaluate a) \(\cos\left(\arcsin \frac{3}{5}\right)\) and b) \(\tan\left(\cos^{–1} \left(–\frac{2}{3}\right)\right)\).

**Solutions**

Start with the inner function, \(\arcsin \frac{3}{5}\). \(\sin x = \frac{opposite}{hypotenuse}\), so draw a right triangle in quadrant 1 and label the acute angle by the origin. The opposite side is 3 and the hypotenuse is 5. See

**Figure 2**. Use the Pythagorean theorem to find the other side.3

^{2}+*b*^{2}= 5^{2}

*b*= 4Now evaluate the outer function, cos, of the angle.

\(\cos u = \frac{adj}{hyp}\)

\(\cos u = \frac{4}{5}\)Start with the inner function, \(\cos^{–1} \left(–\frac{2}{3}\right)\) and draw a right triangle. Since the ratio of sides is negative, draw the triangle in the negative quadrant of the inverse trigonometric function. For cos

^{–1}the negative quadrant is quadrant 2. Label the angle by the origin and the adjacent side –2 and hypotenuse 3. See**Figure 3**. Use the Pythagorean theorem to find the other side.(–2)

^{2}+*b*^{2}= 3^{2}

\(b = \sqrt{5}\)Now evaluate the outer function, tan, of the angle.

\(\tan u = \frac{opp}{adj}\)

\(\tan u = –\frac{\sqrt{5}}{2}\)

Evaluate \(\sin\left(\arctan \left(–\frac{12}{5}\right)\right)\).

\(–\frac{12}{13}\)

Rewrite as an algebraic expression a) sin(arccos(*x*)) and b) tan(sin^{–1}(2*x*)).

**Solutions**

Draw a right triangle and label the sides. The ratio of the sides is \(x = \frac{x}{1}\). Since the inner function is \(\arccos\left(\frac{x}{1}\right)\), the adjacent side is

*x*and the hypotenuse is 1. See**Figure 4**. Use the Pythagorean theorem to find an expression for the third side.*x*^{2}+*b*^{2}= 1^{2}

*b*^{2}= 1 –*x*^{2}

\(b = \sqrt{1 – x^{2}}\)Now evaluate the outer function, sine.

\(\sin u = \frac{opp}{hyp} = \frac{\sqrt{1 – x^{2}}}{1}\)

\(\sin u = \sqrt{1 – x^{2}}\)Draw a right triangle and label the sides. The ratio of the sides is \(2x = \frac{2x}{1}\). Since the inner function is \(\sin^{–1}\left(\frac{2x}{1}\right)\), the opposite side is 2

*x*and the hypotenuse is 1. See**Figure 5**. Use the Pythagorean theorem to find an expression for the third side.*a*^{2}+ (2*x*)^{2}= 1^{2}

*a*^{2}+ 4*x*^{2}= 1

*a*^{2}= 1 – 4*x*^{2}

\(a = \sqrt{1 – 4x^{2}}\)Now evaluate the outer function, tangent.

\(\tan u = \frac{opp}{adj}\)

\(\tan u = \frac{2x}{\sqrt{1 – 4x^{2}}}\)

Rewrite as an algebraic expression: \(\cos\left(\arctan\left(\frac{x}{2}\right)\right)\).

\(\frac{2}{\sqrt{x^{2} + 4}}\)

- tan
^{−1}(sin(*π*)) (6.3#25) - \(\tan^{−1}\left(\sin\left(\frac{π}{3}\right)\right)\) (6.3#27)
- \(\tan^{−1}\left(\sin\left(\frac{4π}{3}\right)\right)\) (6.3#29)
- \(\tan^{−1}\left(\sin\left(−\frac{5π}{2}\right)\right)\) (6.3#31)
- \(\sin\left(\cos^{−1}\left(\frac{3}{5}\right)\right)\) (6.3#33)
- \(\cos\left(\tan^{−1}\left(\frac{12}{5}\right)\right)\) (6.3#35)
- tan(sin
^{−1}(*x*− 1)) (6.3#37) - \(\cos\left(\sin^{−1}\left(\frac{1}{x}\right)\right)\) (6.3#39)
- \(\tan\left(\sin^{−1}\left(x + \frac{1}{2}\right)\right)\) (6.3#41)
- For what value of
*x*does sin*x*= sin^{−1}*x*? Use a graphing calculator to approximate the answer. - (4-08) Evaluate \(\arctan\left(\frac{\sqrt{3}}{3}\right)\) (RW)
- (4-08) Evaluate sin
^{–1}1 (RW) - (4-05) If \(\sin θ = \frac{2}{3}\) and tan
*θ*< 0, find a) cos*θ*and b) cot*θ*. (RW) - (3-04) Solve 2
^{x + 2}= 64 (RW) - (2-09) Solve 2
*x*^{2}+ 3*x*+ 1 > 0 (RW)

For the following exercises, find the exact value, if possible, without a calculator. If it is not possible, explain why.

For the following exercises, find the exact value of the expression in terms of x with the help of a reference triangle.

Mixed Review

- 0
- 0.71
- -0.71
- \(−\frac{π}{4}\)
- \(\frac{4}{5}\)
- \(\frac{5}{13}\)
- \(\frac{x − 1}{\sqrt{−x^2 + 2x}}\)
- \(\frac{\sqrt{x^2−1}}{x}\)
- \(\frac{2x+1}{\sqrt{−4x^2-4x+3}}\)
*x*= 0- \(\frac{π}{6}\)
- \(\frac{π}{2}\)
- \(\frac{\sqrt{5}}{3}\); \(\frac{\sqrt{5}}{2}\)
- 4
- \(\left(−∞, −1\right) ∪ \left(−\frac{1}{2}, ∞\right)\)