FIND sin(t) graph it is a polynomial in sin(t), so there is a polynomial such that:

f(t)=g(sin(t))?

it has horizontal tangent lines at sin(t) = 1/3 or sin t = 1 or sin t = -1 the graph passes through (0,1) and(pi/2 , 4) interval [-pi,pi] FIND THE FINCTION f(t)

(using these clues and derivatives)

f(t) = 1 - 6 sin t + 9 sin^2 t equals to g(sin t), where g(x) = (1 - 3x)^2

Check: f(0) = 1; f(pi/2) = 1 - 6 + 9 = 4

f '(t) = - 6 cos t + 18 sin t cost = 18 cos t (sin t - 1/3)

The tangent to the graph is horizontal when the derivative is 0, that is, if sin t = 1/3 or if cos t = 0 (note that if

sin t = +/- 1, then cos t = 0 by the Pythagorean identity: sin^2 t + cos^2 t = 1).

I am confused about the tangent line part do I plug in the values into the derivative function?

Yes, you just plug the values of sin t into the derivative.

t = pi/2 + pi*k , t = sin^{-1} (1/3) + 2*pi*k and t = -sin^{-1} (1/3) + pi + 2*pi*k (k = 0, +-1, +-2, ...).

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