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If sinx=4/5 and x is in quadrant 2, what is tan2x?
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|dw:1399595270687:dw|Using the Pythagorean Theorem allows us to solve for the missing side. Since we're in the second quadrant, our length along the x-axis is negative.
Recall your Tangent Double Angle Identity:\[\Large\rm \tan2x=\frac{2\tan x}{1-\tan^2x}\]
So just use the triangle to simplify this expression: \(\Large\rm \dfrac{2\tan x}{1-\tan^2x}\) Oh also, in the picture I should called the angle x, not theta. My bad.
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