tan(sin-1(2x))
sine = opposite / hypotenuse tan = opposite / adjacent arcsin is inverse of sin... so sin-1(2x) is the angle of this triangle: |dw:1414804440088:dw| you can use pythag. to get the other side in terms of x. And then the tan of that angle is simply the opposite (2x) over adjacent (what you solved for with pythag.)
thats tan (arcsin(2x))
arcsin 2x means 'the angle whose sine is 2x'
This is how I like to approach these questions: \[\LARGE \tan(\sin^{-1}(2x))\] I start out by noticing that I always take tan, sin, or cos of angles, so I call that thing inside there an angle like this: \[\LARGE \tan( \theta)\] so if you compare these two you can see that you're just finding tangent of an angle. But they're the same look: \[\LARGE \tan(\sin^{-1}(2x))=\tan ( \theta)\] so I set the inside equal too \[\LARGE \sin^{-1}(2x)= \theta\] Now I can solve for this to get: \[\LARGE 2x=\sin(\theta)\] so if I know 2x=sin theta then I can draw it out: |dw:1414898814291:dw| now just use the pythagorean theorem to solve for the other side and then you can solve your original problem tan(theta) by just dividing the opposite by the adjacent side.
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