The Inverse Sine, Cosine and Tangent Functions

Let's review a few things about inverse functions. To have an inverse function, a function must be one-to-one (remember if a horizontal line intersections the graph of a function in more than one place it is NOT one-to-one). If we have points on a function graph and we trade x and y places we'll have points on the inverse function graph. Functions and their inverses "undo" each other so Since x and y trade places, the domain of the function is the range of the inverse and the range of the function is the domain of the inverse The graph of a function and its inverse are reflections about the line y = x (a 45° line).

Is y = sin x a one-to-one function? If we only look at part of the sine graph where the x values go from -/2 to /2 and the y values go from -1 to 1, we could find an inverse function. Is y = sin x a one-to-one function? No! A horizontal line intersects its graph many times. If we want to find an inverse sine function, we can't have the sine repeat itself so we are only going to look at part of the sine graph.

We are going to define the inverse function of sine then to be y = sin-1x. Remember for inverse functions x and y trade places so let's take the values we got for sin x and we'll trade them places for y = sin-1x with our restricted domain. x y = sin x x y = sin-1 x Remember we are only looking at x values from -/2 to /2 so that we can have a one-to-one function that will have an inverse. So for y = sin-1 x, we can put in numbers between -1 and 1. What we get out is the angle between -/2 to /2 that has that sine value.

Let's graph both of these. Here is the graph of sin x between -/2 to /2 x y = sin-1 x x y = sin x Notice they are reflections about a 45° line Here is the graph of sin-1 x from -1 to 1

Since sin x and sin-1 x are inverse functions they "undo" each other. CAUTION!!! You must be careful that the angle is between -/2 to /2 to cancel these.

Looking at the unit circle, if we choose from -/2 to /2 it would be the right half of the circle. It looks like the answer should be 11/6 but remember the range or what you get out must be an angle between -/2 to /2, so we'd use a coterminal angle in this range which is: When you see inverse sine then, it means they'll give you the sine value and it's asking which angle on the right half of the unit circle has this sign value. Notice the y values (sine values) never repeat themselves in this half the circle so the sine function would be 1-to-1 here. This is asking, where on the right half of the unit circle is the sine value -1/2?

Remember when you have sin-1 x it means "What angle between -/2 to /2 has a sine value of x?" Let's look at the unit circle to answer this. This is 5/3 but need This is asking, "What angle from -/2 to /2 has a sine value of square root 2 over 2?" This is asking, "What angle from -/2 to /2 has a sine value of negative square root 3 over 2?"

Now remember for inverse sign we only use values on the right half of the unit circle. So this is asking, "Where on the right half of the circle is the sine value square root 2 over 2?" This looks like sine and its inverse should cancel out but then you'd be getting an answer out that was not in the range so you must be careful on these. Let's work the stuff in the parenthesis first and see what happens.

Let's think about an inverse cosine function Let's think about an inverse cosine function. If we chose the right half of the unit circle for cosine values would we have a one-to-one function? NO! For example: Let's look at the graph of y = cos x and see where it IS one-to-one From 0 to 

So cos-1 x is the inverse function of cos x but the domain is between -1 to 1 and the range is from 0 to . So this is asking where on the upper half of the unit circle does the cosine value equal 1/2.

tan-1 x is the inverse function of tan x but again we must have restrictions to have tan x a one-to-one function. We'll take tan x from -/2 to /2

For help on using your calculator to compute inverse trig functions, click here.