[dispmath]\cos x\cdot\cos 2x\cdots\cos\left(2^nx\right)=\frac{\sin\left(2^{n+1}x\right)}{2^{n+1}\sin x}[/dispmath]

[inlmath]\underline{n=1:}[/inlmath]
[dispmath]\cos x\cdot\cos 2x=\frac{\sin 4x}{4\sin x}=\frac{2\sin 2x\cos 2x}{4\sin x}=\frac{\cancel 4\cancel{\sin x}\cos x\cos 2x}{\cancel 4\cancel{\sin x}}=\cos x\cdot\cos 2x\quad\top[/dispmath]
Indukcijska pretpostavka: [inlmath]\underline{n=k:}[/inlmath]
[dispmath]\cos x\cdot\cos 2x\cdots\cos\left(2^kx\right)=\frac{\sin\left(2^{k+1}x\right)}{2^{k+1}\sin x}[/dispmath]
[inlmath]\underline{n=k+1:}[/inlmath]
[dispmath]\cos x\cdot\cos 2x\cdots\cos\left(2^kx\right)\cos\left(2^{k+1}x\right)=\frac{\sin\left(2^{k+2}x\right)}{2^{k+2}\sin x}[/dispmath]
[dispmath]\cos x\cdot\cos 2x\cdots\cos\left(2^kx\right)\cancel{\cos\left(2^{k+1}x\right)}=\frac{\cancel 2\sin\left(2^{k+1}x\right)\cancel{\cos\left(2^{k+1}x\right)}}{\cancel 2\cdot 2^{k+1}\sin x}[/dispmath]
[dispmath]\cos x\cdot\cos 2x\cdots\cos\left(2^kx\right)=\frac{\sin\left(2^{k+1}x\right)}{2^{k+1}\sin x}\quad\top[/dispmath]