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How do you find maclaurin series for #sinx cosx# using maclaurin for #sinx#?

Answer 1

Cameron Alexander

#sinx cosx = sum_{k=0}^{oo}2^{2k}(-1)^k(x^{2k+1})/((2k+1)!)#

We know #sin(a+b)=sin(a)cos(b)+sin(b)cos(a)# if #a = b# #sin(2a)=2sin(a)cos(a)# then #sinx cosx = sin(2x)/2#

Assuming that

#sin x = sum_{k=0}^{oo}(-1)^k(x^{2k+1})/((2k+1)!)#

then

#sinx cosx = sum_{k=0}^{oo}2^{2k}(-1)^k(x^{2k+1})/((2k+1)!)#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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