Let's take a look at this triangle and try to calculate the lengths of the unknown sides and angles.
Well, the hypotenuse is easy to find using the Pythagorean theorem. It's just
Taking the square root of both sides, we find
But what now? There are no angles. We can easily calculate the sine, cosine and tangent of angle a, they're just
What we need to find those angles are inverse functions, functions that undo the action of the trig functions on both sides of the equation. For example,
You should recall the properties of inverse functions from your study of functions in general, and from learning about the inverse relationship between log and exponential functions. So the inverse relationship between sin(x) and its inverse is:
So let's take that sine relationship from the little table above and use the inverse sine:
The left side is just our angle a
And we find the right side by using the sin-1 button on any scientific calculator:
The inverse trig functions are generally the second [2nd] functions of the trig buttons. Take a look at your own calculator to confirm this.
You should also get in the habit of using the π button instead of typing in 3.14159 ... for π. It's always a good idea to use as much precision in your numbers as possible while you do a calculation, then round the number down to a reasonable number of digits when you report it.
Most calculators don't have sec, csc and cot functions because it's easy enough to type in 1/sin(x), 1/cos(x), or 1/tan(x).
Inverse functions are good for undoing the action of a function on an independent variable. One way to think about inverse functions is that functions always take us from domain to range (i.e. we give them a point in the domain and they give us the corresponding point in the range),
and inverse functions take us from range to domain. In trigonometry, we need inverse functions to solve for the values of angles when all we have are the sides of a triangle.
But because of the periodic nature of the trig functions, we need to be very careful about their definitions and restrictions, and some quirks of their use.
The inverse sine function, sin-1(x), sometimes called the arcsine, arcsin(x), is shown in this figure →
The black curve is the sine function. We know that the graph of its inverse is simply the reflection of the graph of sin(x) across the line y=x, as shown (see functions).
But this presents a problem: The red curve is multiple-valued: It has more than one value for each element of the domain. In order to overcome this, we truncate the domain of the function between -1 and 1, which restricts its range between -π/2 and π/2
This necessary restriction can have some implications on some calculations, and we'll deal with those below.
The case is just about the same for the cos-1(x), or arccos(x) function. In this case we restrict the domain in the same way, but now the range is [0, π].
Again, this will have very real implications for our later calculations. Namely, the inverse sin, cos and tan functions will give, in some cases, answers that need to be adjusted by adding 180˚ or π.
The tan-1(x) function, or arctan(x), is a little different but the basic ideas are the same.
The black curves represent the asymptotic tangent curves with which we're familiar. If we perform our reflection over the line y=x, we get a vertical stack of curves, which are of course multiple-valued; we have to pick one.
We choose, by convention, the central curve, which makes the domain of tan-1(x) [-∞, ∞] and the range [-π/2, π/2].
Now let's look at the problems these domain and range restrictions can have on each of the three main inverse trig. functions.
Because the range of the inverse sine function is [-π/2, π/2], it can never yield an angle greater than π/2 and less than -π/2 (or 3π/2). That's a problem, but not as much as you might think. Study the figure below to get comfortable with the domain and range of sin-1(x).
Find the polar coordinates of the point (-2, 2).
We know this point is in the second quadrant of the graph, and therefore out of the range of the sin-1(x) function. We find tan-1(-2/2) = tan-1(-1) = -45˚,
but that's not where the point is. If we add 180˚ to -45, we get 135˚, the correct angle. You might want to sketch a graph to convince yourself.
Because the range of the inverse cosine function is [0, π], it can never yield an angle greater than π and less than 2π. If a quick graph of the point in question shows that it is in one of the lower quadrants (white region of the circle below), you will have to add 180˚ or π to the output of the cos-1(x) function. Study the figure below to get comfortable with the domain and range of cos-1(x).
The domain of the inverse tangent function is [-∞, ∞], but its range is [-π/2, π/2]. The inverse tangent function has the same range restriction as the sine function, so if you're seeking angles to go with points in the 2nd and 3rd quadrants (white region in the graph below), you'll have to add 180˚ or π rad to the result you get from the tan-1 function.
Find the two missing angles in this right triangle:
For such a problem, you can solve for the missing angles, a and b, using any trig function you'd like.
It's overkill, but just to illustrate the point, here (below) are both angles found using each of the functions sin(x), cos(x) and tan(x).
For efficiency, you'd really want to find the third angle by subtracting the first from 90˚, of course.
For each, we make the appropriate SOH-CAH-TOA expression, such as sin(a) = 5/9.43, then apply the inverse function to both sides (sin-1(x) in this case), to find the angle. The (rounded) solutions are a = 32˚ and b = 58˚, no matter how you get them.
If you aren't yet familiar with the law of cosines, don't worry about this next section. You can find the LOC here.
Find the two missing angles in this non-right triangle:
For a problem like this, we'll use the law of cosines, which contains the term -ab·cos(X), where X is one of the unknown angles.
We solve for cos(X) and find the angle using the inverse cosine function. Here are the law-of-cosines solutions for each of the angles a, b & c.
These solutions are overkill again. It's really only necessary to solve for one angle using the law of cosines. Then the law of sines and the sum of angles of a triangle (180˚) can be used to find the others. Still, it's reassuring when the angles found with only the LOC add to 180˚!
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