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- Feb 5, 2012

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**Original Title: could you help me on this trigonometry function**

Hi rlaydab,rlaydab said:Solve : \(\sin^{2}n\theta - \sin^{2}(n-1)\theta = \sin^{2}\theta\)

\[\sin^{2}n\theta - \sin^{2}(n-1)\theta = \sin^{2}\theta\]

\[\Rightarrow [\sin n\theta - \sin (n-1)\theta][\sin n\theta + \sin (n-1)\theta] = \sin^{2}\theta\]

Using sum to product identities we get,

\[\sin(2n-1)\theta\sin\theta=\sin^{2}\theta\]

\[\Rightarrow \sin\theta[\sin(2n-1)\theta-\sin\theta]=0\]

\[\Rightarrow \sin\theta\sin(n-1)\theta\cos n\theta=0\]

\[\Rightarrow \theta=n_{1}\pi\mbox{ or }(n-1)\theta=n_{2}\pi\mbox{ or }n\theta=2n_{3}\pi\mbox{ where }n_{1},\,n_{2},\,n_{3}\in\mathbb{Z}\]

\[\therefore \theta=n_{1}\pi\mbox{ or }\theta=\frac{n_{2}\pi}{n-1}\mbox{ or }\theta=\frac{2n_{3}\pi}{n}\mbox{ where }n_{1},\,n_{2},\,n_{3}\in\mathbb{Z}\]

Kind Regards,

Sudharaka.