Inverse Functions Study Guide
In this lesson, we introduce inverse trigonometric functions. With these, we are able to find the angles of right triangles whose sides are known.
When we plug an angle θ into a trigonometric function like sine, out comes a ratio r of lengths. The inverse sine (written either sin–l or arcsin) is a function that does the exact opposite. If you plug in a ratio r of sides, out will come the corresponding angle θ.
For example, sin(30°) = , so sin–l() = 30°. Similarly, sin(90°) = 1, so sin–l(l) = 90°. In general, if sin(θ) = r, then sin–l(r) = θ.
Said another way, sin–l(r) = the angle θ for which sin(θ) = r.
The one difficulty that arises is that several angles can have the same ratio, for example, sin(45°) = and sin(l35°) = . What should sin–l be? It cannot output two things. To avoid this problem, it has been decided that only angles –90° ≤ θ ≤ 90° can come out of the sin–l function. Because –90° ≤ 45° ≤ 90° (and 135° is not), we have sin–l = 45°. It might help to see the graph of y = sin(x) on and y = sin–l(x) Figure 14.1.
Notice that the second graph is related to the first by reversing the roles of x and y. In general, the inverse function is found in this way, by reversing the x- and y-axes—in other words, "flipping" the graph about the line y = x.
There are also inverse functions to cosine and tangent. The inverse to cos(θ) = r is cos–l(r) = θ), where the angle θ must be 0° ≤ θ ≤ 180° or 0 ≤ θ ≤ π. The inverse to tan(θ) = r is tan–l(r) = 0, where the angle θ must be –90° < θ < 90° or .
cos–l = 60° = because cos(60°) = and 0 ≤ 60° ≤ 180°
Combining Functions with Inverses
We can plug an inverse trigonometric function into a regular function like tan. This asks, "What is the tangent of the angle whose sine is ?" We have already dealt with such questions before, although now we use new notation.
Remember that sin–1 is the angle θ with sin(θ) = . Although we don't know what θ is exactly, we can draw a right triangle with angle θ, as in Figure 14.2.
The opposite side divided by the hypotenuse must be . The easiest lengths to choose are O = 2 and H = 5. Using the Pythagorean theorem, the adjacent side is A = √25 – 4 = √21. Thus,
tan–1(4) = θ, where tan(θ) = 4
The angle θ is sketched in Figure 14.3.
We choose O = 4 and A = 1 so that . The hypotenuse H = √16 + 1 = √17 is found by the Pythagorean theorem. Thus,
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