2
$\begingroup$

Prove that $\tan{nA}=\frac{\binom{n}{1}\tan{A}-\binom{n}{3}\tan^3{A}+\binom{n}{5}\tan^5{A}-\cdots}{\binom{n}{0}\tan^0{A}-\binom{n}{2}\tan^2{A}+\binom{n}{4}\tan^4{A}-\cdots}$

This is a question from the chapter permutations and combinations and I have no idea as to how to apply those concepts in this question and it would be great if I could get a hint...

  • 0
    Chapter of what? – 2017-01-14
  • 0
    @martycohen Book: *Challenge and Thrill of Pre-college Mathematics* – 2017-01-14
  • 1
    Since this entails the double-angle formula for the tangent function, I guess you will need something that is at least as powerful as addition formulas for the trigonometric functions. ([**De Moivre's formula**](https://en.wikipedia.org/wiki/De_Moivre's_formula) which I hinted in my deleted comment was such an example.) Have you heard of them? – 2017-01-14
  • 0
    @SangchulLee I know the addition formulas and the double-angle formulas but not the De moivre's formula – 2017-01-14
  • 1
    @SangchulLee thank you, got it! – 2017-01-14
  • 0
    @OsheenSachdev, See http://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series/346382#346382 – 2017-01-14

1 Answers 1

2

Hint Perhaps this can help.

$$\left(\frac{1 + i \tan A}{1 - i \tan A}\right)^n=\left(\frac{\cos A + i \sin A}{\cos A - i \sin A}\right)^n=\left(\frac{e^{iA}}{e^{-iA}}\right)^n=\frac{\cos nA + i \sin nA}{\cos nA - i \sin nA}=\frac{1+i \tan nA}{1-i \tan nA}$$

  • 0
    I don't know how to use this formula...seen it for the first time – 2017-01-14
  • 0
    Oh got it...binomial expansion and then componendo and dividendo, right? – 2017-01-14
  • 0
    @OsheenSachdev yes you got it!! – 2017-01-14