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Find an equation of the tangent to the curve at the given point.

y=4(sinx)^2 point: pi/6,1)

So I took the derivative of the original function to get:
y' = 8cosx*sinx

I then chose a point to plug in to find a point for the slope. i picked pi/6 because i thought it would be easier with the cos and the sin but this is as far as I got.
I'm not sure if i've been doing the steps right, or if that's how to find the tangent of a curve.

I know the pointslop form is y-1=m(x+b)

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  1. your derivative will be the slope
    and since (π/6,1) lies on the curve,
    slope of tangent = 8cosπ/6 sinπ/6
    = 8(√3/2)(1/2) = 2√3

    ( you might recall that sin 2x = 2sinxcosx
    so 8sinxcosx = 4(2sinxcosx) = 4 sin2x
    then slope = 4sin(π/3) = 4√3/2 = 2√3 )

    now we have m and a point

    y - 1 = 2√3(x - π/6)
    y -1 = 2√3 x - π/3
    y = 2√3x + 1-π/3

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