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what is the Taylor expansion of the function $\cot(x+(π/4))$? Ignore the powers of $x$ which are bigger than $4$.

I think that there should be other easy ways instead of getting derivatives of $\cot x$ and..... then plugging $x+(π/4)$ ...by the way is there any shortcuts??

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  • $\begingroup$ If you wanted the first 100 terms, I suspect there are shortcuts. But for the first four terms, finding them by brute force may be no more work than shortcuts would be. $\endgroup$
    – Michael Hardy
    Commented Aug 14, 2013 at 18:55
  • $\begingroup$ At what point do you want to get the expansion? $\endgroup$
    – Mhenni Benghorbal
    Commented Aug 14, 2013 at 19:20
  • $\begingroup$ wolframalpha.com/input/?i=Series[Cot[x%2Bpi%2F4]%2C{x%2C0%2C4}] $\endgroup$
    – Adrian Keister
    Commented Aug 14, 2013 at 19:22
  • $\begingroup$ You only need to find the first four derivatives! $\endgroup$
    – Mhenni Benghorbal
    Commented Aug 14, 2013 at 19:40

1 Answer 1

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One alternative is to rewrite $\cot(x+\frac{\pi}{4})$ as $$\frac{\cos(x+\frac{\pi}{4})}{\sin(x+\frac{\pi}{4})}=\frac{\cos x(\frac{\sqrt{2}}{2})-\sin x(\frac{\sqrt{2}}{2})}{\sin x(\frac{\sqrt{2}}{2})-\cos x(\frac{\sqrt{2}}{2})}=\frac{\cos x-\sin x}{\sin x + \cos x},$$ and then multiplying by $\frac{\cos x+\sin x}{\cos x + \sin x}$ to get $$\frac{\cos 2x}{1+\sin 2x}.$$

Then you can divide $\cos 2x=1-2x^2+\frac{2}{3}x^4-\cdots$

by $1+\sin 2x=1+2x-\frac{4}{3}x^3+\cdots$

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