Undoing a function that is not one-to-one
Tonight's idea
An inverse function runs a machine backwards (where you met this). That only works if each output came from one input. Sine fails: \(\sin 30° = \sin 150° = \tfrac12\). "The angle whose sine is \(\tfrac12\)" is not one angle.
The fix is to throw most of the curve away. Keep one piece on which sine takes every value from \(-1\) to 1 exactly once: the climb from \(-90°\) to \(90°\). The inverse of that piece is arcsin, written \(\sin^{-1}\) on a calculator. (The \(-1\) means "inverse", not "one over".)
\(\arcsin x\) is the one angle from \(-90°\) to \(90°\) whose sine is \(x\).
\(\arccos x\) is the one angle from \(0°\) to \(180°\) whose cosine is \(x\).
\(\arctan x\) is the one angle strictly between \(-90°\) and \(90°\) whose tangent is \(x\).
Each window is a choice, made so that no value is hit twice and none is missed. The cost: \(\sin(\arcsin x) = x\) always, but \(\arcsin(\sin\theta) = \theta\) only when \(\theta\) is inside the window. Outside it, you get back a different angle with the same sine.
Watch one: \(\arcsin(\sin 200°)\)
The answer cannot be \(200°\), which is outside the window. Work in two steps.
First the sine. \(200°\) is \(20°\) past \(180°\), in quadrant III, so \(\sin 200° = -\sin 20°\).
Now ask which angle from \(-90°\) to \(90°\) has that sine. Since \(\sin(-\theta) = -\sin\theta\), it is \(-20°\):
\[\arcsin(\sin 200°) = -20°.\]
You have made this kind of choice before. Squaring is not one-to-one either. Keep only \(x \ge 0\) and the inverse is \(\sqrt{x}\), which is why \(\sqrt9\) is 3 and never \(-3\).
Your turn
- Give exact values in degrees, without a calculator: \(\arcsin\tfrac{\sqrt3}{2}\), \(\;\arccos\left(-\tfrac12\right)\), \(\;\arctan(-1)\), \(\;\arcsin(-1)\).
- Find \(\arcsin(\sin 130°)\), \(\;\arccos(\cos 300°)\) and \(\arctan(\tan 135°)\). None of them is the angle you started with.
- \(f(x) = x^2 - 4x + 1\) is not one-to-one. Complete the square, restrict \(f\) to \(x \ge 2\), and find the inverse and its domain. What changes if you keep \(x \le 2\) instead?
- (a) Explain why the window \(-90°\) to \(90°\) would be a bad choice for cosine. Give two separate reasons. (b) Show that \(\cos(\arcsin x) = \sqrt{1 - x^2}\), and say why the answer is never the negative root.
Stretch: two windows that fit together
Try \(\arcsin x + \arccos x\) on a calculator for a few values of \(x\), including a negative one. Guess the rule, then prove it. Start from \(\theta = \arcsin x\) and the fact that \(\cos(90° - \theta) = \sin\theta\). Which window must \(90° - \theta\) lie in?
Answers
- \(60°\), \(120°\), \(-45°\), \(-90°\).
- \(\sin 130° = \sin 50°\), so \(\arcsin(\sin 130°) = 50°\). \(\cos 300° = \cos 60°\), so \(\arccos(\cos 300°) = 60°\). \(\tan 135° = -1\), so \(\arctan(\tan 135°) = -45°\).
- \(f(x) = (x - 2)^2 - 3\). For \(x \ge 2\): \(y + 3 = (x-2)^2\) gives \(x - 2 = +\sqrt{y + 3}\), so \(f^{-1}(x) = 2 + \sqrt{x + 3}\), with domain \(x \ge -3\) (the range of \(f\)). Keeping \(x \le 2\) instead gives \(f^{-1}(x) = 2 - \sqrt{x + 3}\), on the same domain.
- (a) It is not one-to-one there: \(\cos(-60°) = \cos 60°\). And it misses values: cosine is never negative between \(-90°\) and \(90°\), so \(\arccos(-\tfrac12)\) would have no answer. (b) Let \(\theta = \arcsin x\), so \(\sin\theta = x\). Then \(\cos^2\theta = 1 - \sin^2\theta = 1 - x^2\). Because \(\theta\) lies between \(-90°\) and \(90°\), \(\cos\theta \ge 0\), so \(\cos\theta = +\sqrt{1 - x^2}\).
Stretch. The sum is always \(90°\). Let \(\theta = \arcsin x\), so \(-90° \le \theta \le 90°\) and \(\sin\theta = x\). Then \(\cos(90° - \theta) = \sin\theta = x\), and \(90° - \theta\) lies between \(0°\) and \(180°\), which is exactly the arccos window. So \(\arccos x = 90° - \theta\), and \(\arcsin x + \arccos x = 90°\) for every \(x\) from \(-1\) to 1.