I want to prove an identity that involves Inverse Trig Functions. Inverse sine of x plus inverse cosine of x equals pi over 2. Now let's start with an identity we already know, the cofunction identity; sine of pi over 2 minus theta equals cosine of theta.
Now this works for any angle theta, so for example it works for this angle. Remember that inverse trig functions give you an angle. So I can write sine of pi over 2, minus inverse cosine of x equals cosine of inverse cosine of x.
Now let's take a look at this expression. This expression equals x if x is between -1, and 1. Remember that for the inverse angles identity, they're only true if x is within the domain of inverse cosine, in this case inverse cosine is at the domain between -1, 1.
So we have sine of pi over 2, minus inverse cosine of x equals x. Now let's take this over here and we're going to need the definition of inverse sine. I had the sine of pi over 2 minus inverse cosine of x equals x. So I need to turn this into an inverse sine equation, pi over 2 minus inverse cosine of x equals inverse cosine of x. Then all I have to do is add inverse cosine to both sides. Inverse sine of x plus inverse cosine of x equals pi over 2.
This is a nice identity because it shows you the relationship between inverse sine and inverse cosine. If you wanted to know the inverse cosine of an angle of a number, and you know the inverse sign of the number, all you have to do is take pi over 2 and subtract the inverse cosine.