## anonymous 5 years ago Please Help!!! Rewrite the expression as an algebraic expression. sin (tan^-1 (4x)- sin^-1 (4x))

Okay, well you know what tan is right? It is the opposite Length of a Right Triangle, divided by the Adjacent Length based on that angle. $Tan^{-1}()$ is the inverse operation and states that the angle is related to this function. Thus the interior $4x$ must be the supposed "Tan" of an angle. $Tan(Tan^{-1}(x)) = x$ in the domain $-\pi/2<x<\pi/2$ This same idea relates for all the trig functions and their inverses. The domain for which this applies for Sin is $\pi/2\le x\le\pi/2$. For this particular problem you can remember that $sin(\alpha + \beta) = sin(\alpha)cos(\beta) + sin(\beta)cos(\alpha)$So if you imagine a triangle for this problem make two seperate ones for $Tan^{-1}()$ and for $Sin^{-1}()$ then change the value for sin() and cos(). So that instead of $Tan^{-1}$ you have $Sin^{-1}$ or $Cos^{-1}$. And as far as $sin^{-1}()$ goes, you can keep it as sin, but change it for cos. I hope that helps.