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please help me everyone . what is the derivative of: f(t)=20[1-cos^2(t/1000pi)]

Precalculus
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Yes that would be the case if pi was in the numerator, but I think you said pi was in the denominator though?
\[f(t)=20[1-\cos^2(\frac{ t }{ 1000\pi })\] like that ^
\[\frac{ \sin(\frac{ x }{ 500\pi }) }{ 50\pi }\] is this your answer?

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Other answers:

\[f(t) = 20 - 20\cos ^{2}(t/(1000\pi))\] \[f'(t) = 0 - 20 * 2\cos (t/(1000\pi)) * -\sin (t/(1000\pi)) * (1/(1000\pi))\] \[f`(t) = 20/1000\pi * 2\cos(t/1000\pi)*\sin(t/1000\pi)\] Use identity sin (2x) = 2sinxcosx \[f'(t) = 1/50\pi * \sin (2*t/1000\pi)\]
that acually looks right to me. ^
You can also check your answers by punching the equations into wolframalpha.com
beautiful, thank you.
No problem, good luck!

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