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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, 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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