Rlaydab's question from Math Help Forum

[Sudharaka]

Original Title: could you help me on this trigonometry function

Solve : \(\sin^{2}n\theta - \sin^{2}(n-1)\theta = \sin^{2}\theta\)

Hi rlaydab, :)

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

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

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.

 

[jvicens]

Great job on solving the trigonometry function! Your use of the sum to product identities was a smart approach to solving the equation. Your final solutions are also correct and well-explained. Just a small suggestion, it might be helpful to mention that the solutions are in terms of n, as it might not be clear to someone who is not familiar with the notation. Overall, great work and keep up the good problem-solving skills!