The proof proceeds by induction. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$\displaystyle \sin n \theta = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sin \theta       | r = \sin \theta       | c =  }} {{eqn | r = \sin \theta \binom {1 - 1 } 0 \paren {2 \cos \theta}^{1 - \paren {0 + 1} }       | c = Zero Choose Zero, {{Defof|Integer Power}} }} {{end-eqn}} So $\map P 1$ is seen to hold. $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = \sin 2 \theta       | r = 2 \sin \theta \cos \theta       | c = Double Angle Formulas/Sine }} {{eqn | r = \sin \theta \binom {2 - 1 } 0 \paren {2 \cos \theta}^{2 - \paren {0 + 1} }       | c =  }} {{end-eqn}} So $\map P 2$ is also seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P n$ is true, where $n > 2$, then it logically follows that $\map P {n + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \map \sin {n \theta} = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$ from which we are to show: :$\displaystyle \map \sin {\paren {n + 1} \theta} = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n + 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n + 1 - \paren {2 k + 1} } }$ === Induction Step === This is our induction step: To proceed, we will require the following lemma: === Lemma === {{:Sine of Integer Multiple of Argument/Formulation 1/Lemma}}{{qed|lemma}} {{begin-eqn}} {{eqn | l = \map \sin {\paren {n + 1} \theta}        | r = \map \sin {n \theta + \theta }       | c = }} {{eqn | r = \sin n \theta \cos \theta + \cos n \theta \sin \theta       | c = Sine of Sum }} {{eqn | r = \sin n \theta \cos \theta + \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1 } \theta}        | c = Lemma above }} {{eqn | r = \sin n \theta \paren {2 \cos \theta } - \map \sin {\paren {n - 1 } \theta}        | c =  }} {{eqn | r = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } } \paren {2 \cos \theta } - \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n - 1 - \paren {2 k + 1} } }       | c =  }} {{eqn | r = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k} } } - \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 2} } k \paren {2 \cos \theta}^{n - \paren {2 k + 2} } }       | c =  }} {{eqn | r = \sin \theta \paren {\paren {2 \cos \theta}^n + \paren {\sum_{k \mathop \ge 1} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k} } } + \paren {\sum_{k \mathop \ge 1} \paren {-1}^k \binom {n - \paren {k - 1 + 2} } {k - 1 } \paren {2 \cos \theta}^{n - \paren {2 \paren {k - 1 } + 2} } } }       | c =  }} {{eqn | r = \sin \theta \paren {\paren {2 \cos \theta}^n + \paren {\sum_{k \mathop \ge 1} \paren {-1}^k \paren {\binom {n - \paren {k + 1} } k + \binom {n - \paren {k + 1} } {k - 1 } } \paren {2 \cos \theta}^{n - \paren {2 k} } } }       | c =  }} {{eqn | r = \sin \theta \paren {\paren {2 \cos \theta}^n + \sum_{k \mathop \ge 1} \paren {-1}^k \binom {n - k } k \paren {2 \cos \theta}^{n - \paren {2 k } } }       | c = Pascal's Rule }} {{eqn | r = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - k } k \paren {2 \cos \theta}^{n - \paren {2 k } } }       | c =  }} {{eqn | r = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n + 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n + 1 - \paren {2 k + 1} } }       | c = adding and subtracting $1$ }} {{end-eqn}} The result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{>0}: \sin n \theta = \sin \theta \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }$ {{qed}}