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Using \(\sin^2\theta+\cos^2\theta=1\), the harmonic magnitude simplifies to

Cn=Vnπ2−2cos⁡(2πnD)C_n=\frac{V}{n\pi}\sqrt{2-2\cos(2\pi nD)}

Using the identity \(1-\cos(2x)=2\sin^2x\),

Cn=2Vnπ|sin⁡(πnD)|C_n=\frac{2V}{n\pi}\left|\sin(\pi nD)\right|

Therefore, the magnitude of the \(n\)th harmonic is

Cn=2Vnπ|sin⁡(πnD)|\boxed{C_n=\frac{2V}{n\pi}\left|\sin(\pi nD)\right|}

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