Trigonometry – Finding the Value of n in the Given Identity

trigonometry

Problem :

Given that $\sin^3 x \sin 3x = \sum^n_{m=0}C_m \cos mx, C_n \neq 0 $ is an identity. Find the value of n.

I tried : $\sin3x = 3\sin x – 4\sin^3 x$ but unable to reach to any point…. Please suggest further ….Thanks..

Best Answer

Using the identity $$ \sin^3(x)=\frac{3\sin(x)-\sin(3x)}{4} $$ and the identity $$ \sin(ax)\sin(bx)=\frac12\big(\cos((a-b)x)-\cos((a+b)x)\big) $$ we get $$ \begin{align} \sin^3(x)\sin(3x) &=\frac{3\sin(x)-\sin(3x)}{4}\sin(3x)\\ &=\frac38\big(\cos(2x)-\cos(4x)\big)-\frac18\big(\cos(0x)-\cos(6x)\big)\\ &=\frac18\big(\cos(6x)-3\cos(4x)+3\cos(2x)-\cos(0x)\big) \end{align} $$ Thus, $n=6$.

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