>> [music] People often talk about inverse trig functions.
But that's nonsense. The trig functions aren't invertible.
Look at y equals sine x. Sine sends different input values, like 0,
pi, 2 pi, 3 pi, to the same output value of 0.
So, how are you going to pick the inverse for 0?
To get around this problem, we're only going to talk about the inverses of trig
functions after we restrict their domain. Here I've got a graph of sine.
And you can see this function is not invertible.
It fails the horizontal line test. But, if I restrict the domain of sine to
just be between minus pi over 2, and pie over 2, this function is invertible.
And, here's its inverse, arc sine. By convention, arc sine outputs angles
between minus pi over 2 and pi over 2. And also by convention, arc cosine outputs
angles between 0 and pi. And arc tan outputs angles between minus
pi over 2 and pi over 2. Note that I'm calling these inverse trig
functions arc whatever. You know, arc cosine, arc sine.
One reason to call these things arc cosine or arc whatever, is because of radian
measure. If this is a unit circle, then the length
of this arc is the same as the measure of this angle, in radians.
That's the definition of radians. So, to say that theta is arc cosine 1
half, is just to say that theta is the length of the arc whose cosine is 1 half.
Now, what happens when we compose the inverse trig functions with trig
functions? Before complicating matters by thinking
about trig functions, let's think back to an easier example, the square root and the
squaring function. And just like trig functions are not
actually invertible, the squaring function's not invertible, because
multiple inputs yield the same output. So we had to define the square root to
pick the non-negative square root of x. And that means that if x is bigger than or
equal to zero, then I have the square root of x squared is equal to x.
But if x is negative, this is not true. The same sort of deal happens with trig
functions. So if theta is between minus pi over 2 and
pi over 2, then acrc sine of sin of theta is equal to theta.
But if theta's outside of this range, this is not the case.
If theta's ouside of this range, yes, arc sine is going to produce for me, another
angle, whose sine is the same as the angle theta.
But there's plenty of other inputs to sine which yield the same output.
Things are even more complicated if you mix together different kinds of trig
functions. For example, the sine of arc tan of x
happens to be equal to x divided by the square root of 1 plus x squared.
Where would you possibly get a formula like that?
Well the trick is to draw a right triangle, the correct triangle.
Well here I've drawn a right triangle. And I've got an angle theta.
And I've drawn this triangle so that theta's tangent is x.
That means that theta is the arc tangent of x.
Now by the Pythagorean theorem, I know the hypotenuse of this triangle has length of
square root of 1 plus x squared. And that gives me enough information to
compute the sine of theta. The sine of theta is this opposite side x,
divided by the hypotenuse, the square root of 1 plus x squared.
This tells me that the sine of arc tan of x must be x over the square root of 1 plus
x squared. Drawing pictures will get you very far in
understanding all of these relationships. I think it's basically impossible just to
memorize all of the possible formulas that relate inverse trig and trig functions.
But if you're ever wondering what those formulas are, you just draw the
appropriate picture, and then you can figure it out right away.
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