Composing trig and inverse trig functions

When you reach for it

An inverse trig function hands back an angle. Very often that angle is not wanted for itself: it is fed straight into another trig function, as in \displaystyle \cos\left(\arcsin\frac{5}{13}\right). This is a composition, the output of \arcsin becoming the input of \cos. A calculator can grind out a decimal, but the exact value, \displaystyle \frac{12}{13}, comes from a picture, and when the inside holds a variable, as in \displaystyle \tan\left(\arccos\frac{x}{3}\right), only the picture works. The result is then an algebraic expression in x, with no trig left in it.

You reach for this move whenever a trig function is applied to an inverse trig function of a different kind, when a formula such as \cos(\arcsin x) needs to be written without trig, and later, in calculus, wherever an angle was introduced only to be removed again. It rests on the slices chosen in the inverse trig functions: \arcsin returns an angle in \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \arccos one in [0, \pi], and \arctan one in \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).

The procedure

Name the inner angle. Write \theta for the inner expression. In \displaystyle \cos\left(\arcsin\frac{5}{13}\right), let \displaystyle \theta = \arcsin\frac{5}{13}. By the definition of arcsine this says two things: \displaystyle \sin\theta = \frac{5}{13}, and \theta lies in \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. The question has become “find \cos\theta”, and both facts are needed to answer it.

Put a point on the terminal side. Place \theta in standard position and pick a point (a, b) on its terminal side, at distance \displaystyle r = \sqrt{a^2 + b^2} from the origin. Then

The inner function fixes two of a, b, r, and the Pythagorean relation a^2 + b^2 = r^2 gives the third up to a sign. The slice settles the sign, and this is the step that makes the answer exact rather than “plus or minus”.

• \displaystyle \theta = \arcsin\frac{p}{q} ends in the right half of the plane. Take b = p and r = q; then \displaystyle a = \sqrt{q^2 - p^2}, never negative.
• \displaystyle \theta = \arccos\frac{p}{q} ends in the upper half of the plane. Take a = p and r = q; then \displaystyle b = \sqrt{q^2 - p^2}, never negative.
• \displaystyle \theta = \arctan\frac{p}{q}, written with q > 0, ends in the right half of the plane, off the vertical axis. Take a = q and b = p; then \displaystyle r = \sqrt{p^2 + q^2}.

The sign of the input is carried by the known coordinate; the third coordinate is always the positive root, because each slice lies in a half plane where that coordinate is never negative. When the input is positive, \theta is acute and the point (a, b) is simply the corner of a right triangle with legs a, b and hypotenuse r: the reference triangle.

Read off the outer function. With a, b and r in hand, the outer function is one of the ratios above (or a reciprocal: \displaystyle \csc\theta = \frac{r}{b}, \displaystyle \sec\theta = \frac{r}{a}, \displaystyle \cot\theta = \frac{a}{b}).

Find the domain when the input is a variable. The composition is defined where the inner function accepts its input and where the outer ratio has no zero denominator. For \displaystyle \tan\left(\arccos\frac{x}{3}\right) the first condition is \displaystyle -1 \le \frac{x}{3} \le 1, and the second removes the x that makes a = 0.

The same function inside and out. When the inner and outer functions match, no triangle is needed. Since \arcsin x is an angle whose sine is x, \sin(\arcsin x) = x for every x in [-1, 1]; likewise \cos(\arccos x) = x on [-1, 1] and \tan(\arctan x) = x for every real x. In the other order the inverse can only return an angle from its slice, so \arcsin(\sin y) = y only for y in \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], and outside the slice the answer is the slice angle with the same sine. The prerequisite lesson works through those cases; here they appear mixed with other functions.

Worked examples

Boundaries

• The inner input must be in the inverse’s domain. \cos(\arcsin 2) is undefined, because no angle has sine 2. For a variable input this is the first condition on the domain, as in Example 4.
• The outer function may be undefined at the angle. \displaystyle \tan(\arccos 0) = \tan\frac{\pi}{2} does not exist, and \cot(\arctan 0) = \cot 0 does not exist. Every zero denominator in the final expression marks such a point and is removed from the domain.
• The positive root is right only because of the slice. The rule “third coordinate is \displaystyle +\sqrt{\phantom{x}}” comes from the half plane that each slice occupies. For an angle that was not produced by an inverse function, knowing \displaystyle \sin\theta = \frac{5}{13} leaves \displaystyle \cos\theta = \pm\frac{12}{13}.
• Undoing works in one order only. \sin(\arcsin x) = x on all of [-1, 1], but \arcsin(\sin y) = y only for y in \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]; the same holds for cosine on [0, \pi] and tangent on \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).
• An identity between inverses holds only where the angles share a slice. Example 7 holds on (-1, 1); the analogous formula for arccosine fails for negative x (Exercise 24).

Where this goes next

Turning \displaystyle \tan\left(\arccos\frac{x}{3}\right) into \displaystyle \frac{\sqrt{9 - x^2}}{x} is a small case of simplifying trig expressions: an angle is introduced and then removed with the Pythagorean relation, which is the Pythagorean identity in triangle form. The same move, read in reverse, is trigonometric substitution in calculus, where \displaystyle \sqrt{9 - x^2} is simplified by setting x = 3\cos\theta and the answer is translated back to x with exactly the triangle of Example 4. It is also how the derivatives of the inverse trig functions lose all their trig: the derivative of \arcsin x comes out as \displaystyle \frac{1}{\cos(\arcsin x)}, which is \displaystyle \frac{1}{\sqrt{1 - x^2}}.

Exercises

Exact values, positive inputs

Find the exact value.

1. \displaystyle \sin\left(\arccos\frac{8}{17}\right)
2. \displaystyle \tan\left(\arcsin\frac{20}{29}\right)
3. \displaystyle \cos\left(\arctan\frac{7}{24}\right)
4. \displaystyle \sec\left(\arcsin\frac{1}{3}\right)

Exact values, the sign from the slice

Find the exact value.

5. \displaystyle \cos\left(\arcsin\left(-\frac{9}{41}\right)\right)
6. \displaystyle \tan\left(\arccos\left(-\frac{12}{13}\right)\right)
7. \displaystyle \sin\left(\arctan\left(-\frac{3}{4}\right)\right)
8. \displaystyle \sin\left(\arccos\left(-\frac{1}{4}\right)\right)
9. \displaystyle \cot\left(\arcsin\left(-\frac{2}{5}\right)\right)

Mixed and undoing compositions

Find the exact value.

10. \cos(\arccos(-0.35))
11. \displaystyle \arccos\left(\cos\left(-\frac{\pi}{7}\right)\right)
12. \displaystyle \arcsin\left(\cos\frac{5\pi}{6}\right)
13. \displaystyle \arctan\left(\sin\frac{3\pi}{2}\right)
14. \arcsin(\sin 5), also to four decimal places.
15. \displaystyle \cos\left(\arctan\left(\sin\frac{\pi}{2}\right)\right)

Algebraic expressions

Write each as an algebraic expression in x and give its domain.

16. \displaystyle \sin\left(\arccos\frac{x}{5}\right)
17. \cos(\arctan(3x))
18. \displaystyle \tan\left(\arcsin\frac{x}{2}\right)
19. \sec(\arctan x)
20. \sin(\arccos(x + 1))
21. \displaystyle \cot\left(\arccos\frac{x}{4}\right)

Reasoning

22. Explain why \cos(\arcsin x) is never negative, while \cos(\arccos x) is negative for every negative x in [-1, 1].
23. Otieno claims that \sin(\arccos x) and \cos(\arcsin x) are the same expression. Is he right? Give two reasons, one from triangles and one from the fact that \displaystyle \arcsin x + \arccos x = \frac{\pi}{2}.
24. Show that \displaystyle \arccos x = \arctan\frac{\sqrt{1 - x^2}}{x} for 0 < x \le 1 but fails for -1 \le x < 0, and give a correct formula for -1 \le x < 0.
25. Kipchoge says \sin(\arctan x) gets as close to 1 as you like but never equals it. Prove this from the algebraic expression for \sin(\arctan x), and explain it from the arctangent slice.