Maclaurin series

Can someone explain how to do this question: ‘use known Maclaurin series to find the first 3 non-zero terms for the expansion of ln(1+ sinx)’
I thought using known series meant to use the expansion for ln(1+x) so I did that and substituted sinx for x in that expansion

That doesnt work -
https://www.wolframalpha.com/input?i=series++ln%281%2Bx%29
https://www.wolframalpha.com/input?i=series+ln%281%2Bsin%28x%29%29

But the maclaurin is based on finding the functions value and its derivatives at x=0 and subbing that into the series. So just do that. Why do you think theyre different (based on the definition), which is the first term which is different, why (thinking about the local approximations of sin(x))?

Is doing the expansion from scratch by differentiating and evaluating at 0 what is meant by “use known maclaurin series” for this case?
Not sure exactly what you mean but the expansion for sin is completely different. Only odd powers of x are in the expansion and you divide by factorials.

If youve got to do the question the hard way to understand it, then just do it.
But for the actual question, you have the composition of two functions ln() and sin(). The 1+ part isnt that important. So you could use known maclaurin series for either / both to get the series for their composition.
It helps to see some form of attempt.

If I’ve got both of these how would I combine them

Wbf, you seem to be asking "how do I get the answer" without really understanding what youre doing. Did you do it from first principles? What was your working that you referred to in the OP - so what series did you end up with? You know the answer from the previous wolfram link, so have you ended up with anything even if its wrong?

I do understand what I’m doing and I know how to get the answer.
Unsure about the ‘shortcut’ of doing it by using known series

ok, I agree with the "fundamental" way and if youve done it, its good to post your working as youve done here.
If you had to get the series for
ln(1 + x^2)
ln(1 + x + x^2)
what would you do (shortcut/substitution or fundamental if really necessary).

Yes. So you know the series for sin(x) so
ln(1 + sin(x)) = ln(1 + ....) = .....

That’s what I did initially. For every x in the ln(1+x) expansion, I substituted a sinx.
But I ended up with ln(1 + sinx) = sinx - sin^2(x)/2 + sin^3(x)/3 - …
which isn’t the same answer as the ‘fundamental’ way of doing it

You missed the hint about knowing the series for sin(x). So
ln(1 + sin(x)) = lnl(1 + ...)
so what replaces the ... then just do as in your previous post.

Ahh right thank you