if m,n are integers satisfying
1+cos2x+cos4x+cos6x+cos8x+cos10x=cos mx.sin nxsinx
then (m+n) equals ?
A)9 B)10 C)11 D)12
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1 Answers
1708\hspace{-16}$Let $\bf{\mathbb{S}=1+\cos(2x)+\cos(4x)+\cos(6x)+\cos(8x)+\cos(10x)}$\\\\\\ Multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\bf{2.\sin (x)}$, We Get\\\\\\ $\bf{\mathbb{S}=\frac{1}{2.\sin (x)}\left\{2\cos(0).\sin (x)+2\cos(2x).\sin (x)+2\cos(4x).\sin (x)+2\cos(6x).\sin (x)+2\cos(8x).\sin (x)+2\cos(10x).\sin (x)\right\}}$\\\\\\ $\bf{\mathbb{S}=\frac{1}{2.\sin (x)}\left\{\sin(x)-\sin(-x)+\sin(3x)-\sin(x)+\sin(5x)-\sin(3x)+\sin(7x)-\sin(5x)+\sin(9x)-\sin(7x)+\sin(11x)-\sin(9x)\right\}}$\\\\\\ $\bf{\mathbb{S}=\frac{1}{2\sin(x)}\left\{\sin(11x)+\sin(x)\right\}=\frac{1}{2\sin(x)}\times 2\sin(6x).\cos (5x)}$\\\\\\ So $\bf{\mathbb{S}=\frac{\sin (6x).\cos (5x)}{\sin(x)}=\frac{\sin(mx).\cos(nx)}{\sin x}}$(Given)\\\\\\ So $\bf{\left(m\;,n\right)=(6\;,5)\Leftrightarrow m+n = 6+5=11}$
