Proof by induction: For all $n \in \N_{\ge0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \tan \left({n \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\left\lfloor{\frac {n - 1} 2}\right\rfloor} \left({-1}\right)^i \binom n {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\left\lfloor{\frac n 2}\right\rfloor} \left({-1}\right)^i \binom n {2 i} \tan^{2 i}\theta}$ === Basis for the Induction === $P(0)$ is the case: {{begin-eqn}} {{eqn | o =        | r = \frac {\displaystyle \sum_{i \mathop = 0}^{-1} \left({-1}\right)^i \binom 0 {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^0 \left({-1}\right)^i \binom 0 {2 i} \tan^{2 i}\theta}       | c =  }} {{eqn | r = \frac 0 {\displaystyle \sum_{i \mathop = 0}^0 \left({-1}\right)^i \binom 0 {2 i} \tan^{2 i}\theta}       | c = Because the upper bound is smaller than the lower bound, this results in a vacuous summation and thus is zero. }} {{eqn | r = \frac 0 {\left({-1}\right)^0 \dbinom 0 0 \tan^0 \theta}       | c =  }} {{eqn | r = \frac 0 1       | c =  }} {{eqn | r = 0       | c =  }} {{eqn | r = \tan \left({0 \theta}\right)       | c =  }} {{end-eqn}} and so can be seen to hold. $P(1)$ is the case: {{begin-eqn}} {{eqn | o =        | r = \frac {\displaystyle \sum_{i \mathop = 0}^0 \left({-1}\right)^i \binom 1 {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^0 \left({-1}\right)^i \binom 1 {2 i} \tan^{2 i}\theta}       | c =  }} {{eqn | r = \frac {\left({-1}\right)^0 \dbinom 1 1 \tan^1 \theta} {\left({-1}\right)^0 \dbinom 1 0 \tan^0 \theta}       | c =  }} {{eqn | r = \frac {\tan \theta} 1       | c =  }} {{eqn | r = \tan \theta       | c =  }} {{end-eqn}} and so is also seen to hold. These two cases together form the basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k-2}\right)$ and $P \left({k-1}\right)$ is true, where $k > 2$ is an even number, then it logically follows that $P \left({k}\right)$ and $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \tan \left({\left({k - 2}\right) \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta}$ :$\displaystyle \tan \left({\left({k - 1}\right) \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 1} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 1} {2 i} \tan^{2 i}\theta}$ Then we need to show: :$\displaystyle \tan \left({k \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom k {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2} \left({-1}\right)^i \binom k {2 i} \tan^{2 i}\theta}$ :$\displaystyle \tan \left({\left({k + 1}\right) \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2} \left({-1}\right)^i \binom {k + 1} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2} \left({-1}\right)^i \binom {k + 1} {2 i} \tan^{2 i}\theta}$ === Induction Step === This is our induction step: For the first part: {{begin-eqn}} {{eqn | l = \tan \left({k \theta}\right)       | r = \tan \left({\left({k - 2}\right) \theta + 2 \theta}\right)       | c =  }} {{eqn | r = \frac {\tan \left({\left({k - 2}\right) \theta}\right) + \tan \left({2 \theta}\right)} {1 - \tan \left({\left({k - 2}\right) \theta}\right) \tan \left({2 \theta}\right)}       | c = Tangent of Sum }} {{eqn | r = \frac {\frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta} + \tan \left({2 \theta}\right)} {1 - \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta} \tan \left({2 \theta}\right)}       | c = By Induction Hypothesis }} {{eqn | r = \frac {\frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta} + \frac {\displaystyle 2 \tan \theta} {\displaystyle 1 - \tan^2 \theta} } {1 - \frac {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta} \frac {\displaystyle 2 \tan \theta} {\displaystyle 1 - \tan^2 \theta} }       | c = By Double Angle Formula for Tangent }} {{eqn | r = \frac {\left({\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta}\right) \left({\displaystyle 1 - \tan^2 \theta}\right) - \left({\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta}\right) \left({\displaystyle 2 \tan \theta}\right)} {\left({\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 1} \left({-1}\right)^i \binom {k - 2} {2 i} \tan^{2 i}\theta}\right) \left({\displaystyle 1 - \tan^2 \theta}\right) - \left({\displaystyle \sum_{i \mathop = 0}^{\frac k 2 - 2} \left({-1}\right)^i \binom {k - 2} {2 i + 1} \tan^{2 i + 1}\theta}\right) \left({\displaystyle 2 \tan \theta}\right)}       | c =  }} {{end-eqn}} For the second part: {{finish}} So $P \left({k-2}\right) \land P \left({k-1}\right) \implies P \left({k}\right) \land P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N: \tan \left({n \theta}\right) = \frac {\displaystyle \sum_{i \mathop = 0}^{\left\lfloor{\frac {n - 1} 2}\right\rfloor} \left({-1}\right)^i \binom n {2 i + 1} \tan^{2 i + 1}\theta} {\displaystyle \sum_{i \mathop = 0}^{\left\lfloor{\frac n 2}\right\rfloor} \left({-1}\right)^i \binom n {2 i} \tan^{2 i}\theta}$ {{qed}}