En Euler-omskrivning visar att

\begin{array}{l} \sin x \cdot \sin 3 x \cdot \sin 5 x = \left(\frac{e^{ix} - e^{-ix}}{2i}\right) \cdot \left(\frac{e^{i3x} - e^{-i3x}}{2i}\right) \cdot \left(\frac{e^{i5x} - e^{-i5x}}{2i}\right) = \\ - \frac{1}{8i} \left(e^{i4x} - e^{-i2x} - e^{i2x} + e^{-i4x}\right) \left(e^{i5x} - e^{-i5x}\right) = \\ - \frac{1}{8i} \left(e^{i9x} - e^{-ix} - e^{i3x} + e^{-i7x} - e^{i7x} + e^{-i3x} + e^{ix} - e^{-i9x}\right) = \\ \frac{1}{4} \left(\frac{e^{i3x} - e^{-i3x}}{2i} + \frac{e^{i7x} - e^{-i7x}}{2i} - \frac{e^{ix} - e^{-ix}}{2i} - \frac{e^{i9x} - e^{-i9x}}{2i}\right) = \\ \frac{1}{4} (\sin 3x + \sin 7x - \sin x - \sin 9x). \end{array}

Ur detta fås att

4 \sin x \sin 3 x \sin 5 x = \sin 7 x - \sin 9 x \iff \sin 3 x + \sin 7 x - \sin x - \sin 9 x = \sin 7 x - \sin 9 x \iff \sin 3 x = \sin x \iff \left\{ \begin{array}{c} 3 x = x + 2 \pi n \\ \text{eller} \\ 3 x = \pi - x + 2 \pi n \end{array} \right. \iff \left\{ \begin{array}{c} x = \pi n \\ \text{eller} \\ x = \frac{\pi}{4} + \frac{\pi n}{2} \end{array} \right. \quad \text{där } n \in \mathbf{Z}.

Svar: $\sin x \cdot \sin 3x \cdot \sin 5x = \frac{1}{4} (\sin 3x + \sin 7x - \sin x - \sin 9x)$ ,
Ekvationen har lösningarna $\left{

\begin{array}{c} x = \pi n \\ \text{eller} \\ x = \frac{\pi}{4} + \frac{\pi n}{2} \end{array}

\right.$ där $n \in \mathbf{Z}$ .