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  • 5 years ago

Please Help!!! Rewrite the expression as an algebraic expression. sin (tan^-1 (4x)- sin^-1 (4x))

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  1. anonymous
    • 5 years ago
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    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.

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