Description
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)
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Explanation & Answer
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).
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