Taylor Series for sinx about pi/6

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SUMMARY

The Taylor Series for the function f(x) = sin(x) about the center point c = π/6 can be derived using the formula pn(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)²/2! + f'''(c)(x-c)³/3! + ... The calculated derivatives at π/6 yield f(π/6) = 1/2, f'(π/6) = √3/2, f''(π/6) = -1/2, and f'''(π/6) = -√3/2, with the coefficients repeating thereafter. The challenge lies in correctly applying the alternating signs and ensuring that √3 appears only in the odd terms of the series.

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Homework Statement


Determine the Taylor Series for f(x)=sinx about the center point c=pi/6

Homework Equations


pn(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)^2/2! + f'''(c)(x-c)^3/3! + ...

The Attempt at a Solution



f(pi/6) = 1/2
f'(pi/6) = \sqrt{3}/2
f''(pi/6) = -1/2
f'''(pi/6) = -\sqrt{3}/2
f(4)(pi/6) = 1/2 and the coefficients repeats from hereWhat I am stuck on is.. how do I make it so that (-1) will only appear on the 3rd and 4th term... and how do I make it so that \sqrt{3} will only appear on the odd terms?
 
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If w=x-pi/6 then sinx=sin(w+pi/6)=sin(w)cos(pi/6)+sin(pi/6)cos(w). Try expanding this and see what you get
 
I am not sure how that relates to Taylor's Series?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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