Another easy way to find the series for ln(1-sin x) is to take the ln(1-x) series (easily derived from integrating the common geometric series term by term) and substitute sin x for x. That way's actually the fastest method to get any ln(1 +/- something in terms of x) into a Maclaurin series expansion.
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sin(-x) = -sin(x)
Excellent! Thank you.
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π/4 = 1 - 1/3 + 1/5 - 1/7 + ... + (-1)n/(2n + 1) + ...
π2/6 = 1 + 1/4 + 1/9 + 1/16 + ... + 1/n2 + ...
ln(2) = 1 - 1/2 + 1/3 - 1/4 + ... + (-1)n-1/n + ...
not make you happy?
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That way's actually the fastest method to get any ln(1 +/- something in terms of x) into a Maclaurin series expansion.
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