Question

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)

Answer 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