Using the Inverse Trigonometric Functions - Concept 19,936 views
In a problem where two trig functions are not inverses of each other (also known as "inverse trigonometric functions", (1) replace the inverse function with a variable (which represents an angle), (2) use the definition of the inverse function to draw the angle in the unit circle and identify one coordinate, (3) find the missing coordinate (use Pythagorean Theorem, for example), (4) use the coordinates to find the missing value.
When you're studying the inverse trig functions you might come across a problem like this, evaluate tangent of inverse sine of four fifths. Notice these are not inverses of one another so you can't use the inverse identities here. So what I suggest you do is you make a little substitution let's call this, now remember inverse trig functions will give you an angle their output is always going to be an angle. Let's call this alpha, in Math we have a tendency to label angles with Greek letters so let's call it alpha. And I'm going to draw alpha on the unit circle in a second. Let me just use the definition of inverse sine to figure out what kind of an angle alpha is.
Remember the definition of inverse sine y equals the inverse sine of x means x equals sine of y for y between negative pi over 2 and pi over 2. So alpha equals inverse sine of four fifths means sine of alpha equals four fifths and alpha is between negative pi over 2 and pi over 2. And that suggests that alpha is going to be in the first quadrant because we have a positive value. So let me draw alpha like this, so I've drawn alpha in the first quadrant, the sine value is four fifths and in order to solve this problem I'm going to have to figure out what the x coordinate would be. But remember this is a point in the unit circle and on the unit circle x squared plus y squared equals 1. So in this case we'd have x squared plus four fifths squared equals 1 or x squared plus 16 over 25 equals and I'll change this to 25 over 25 and then I subtract 16 over 25 from both sides and I get x squared equals 9 over 25. So x is going to be plus or minus 3 over 5, now which is it?
Well we're in the first quadrant x has got to be positive, so x is three fifths, now that doesn't exactly help me figure out what the tangent of alpha is because remember that's what we're looking for, the tangent of this angle alpha. If I were looking for the cosine o alpha I'd be done it would be three fifths. But remember that the tangent of alpha is the y coordinate divided by the x coordinate. So the tangent of alpha is going to be the y coordinate of four fifths divided by the x coordinate of three fifths and that's going to give me four thirds and that's it so always make a substitution like this. Let's see another example, sine of arc cosine of negative two thirds remember that arc cosine is the same as inverse cosine. So sine of arc cosine of two thirds this is going to be another angle, let me call it beta and I should plot beta in a second but first let's figure out what kind of an angle beta is remember that beta equals inverse cosine of negative two thirds means the cosine of beta equals negative two thirds.
And beta is between 0 and pi remember that, that's the going to be the range of inverse cosine. So if I draw beta on this unit circle I've got to draw beta between 0 and pi somewhere up here. Now where would the cosine be negative? Remember that cosine comes from the first coordinate at the point of the unit circle. It's going to be somewhere in this quadrant, so let me pick a point where it looks like the x coordinate is negative two thirds how about there? Negative two thirds, something and this would be my angle. Now all I have to do is figure out what the y coordinate is, well I still have the fact that x squared plus y squared equals 1 and I can plug in negative two thirds for x. Our negative two thirds squared is0over 9 and I can change the 1 to 9 over 9 and thus if I subtract both sides 4 9 is from both sides I get y squared equals 5 over 9 and y equals plus or minus root 5 over 3.
And you can tell that since we're in the second quadrant the y coordinate is going to be positive. So I should choose y equals plus root 5 over 3. Now I'm looking for the sine of beta, the sine of beta is exactly the y value. So I've got my answer sine of beta which is root 5 over 3 and I'm done. Don't forget this trick of renaming the arc cosine or the inverse trigonometric function value because they always give you angles. You can plot that angle on the unit circle figure out what both coordinates are and then use that to find the sine, cosine or tangent of the result.