The value of the determinant left| {begin{arr | vedclass

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(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}1&1&1\{\cos nx}&{\cos (n + 1)x}&{\cos (n + 2)x}\{\sin nx}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array

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(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}1&1&1\{\cos nx}&{\cos (n + 1)x}&{\cos (n + 2)x}\{\sin nx}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}} ight|$.Applying the column transformation ${C_3} 	o {C_3} - (2\cos x){C_2} + {C_1}$ is not ideal,so let us use the property $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$ and $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$.Applying ${C_1} 	o {C_1} + {C_3}$:$\Delta = \left| {\begin{array}{*{20}{c}}{1 + \cos (n + 2)x}&1&{\cos (n + 2)x}\{\cos nx + \cos (n + 2)x}&{\cos (n + 1)x}&{\cos (n + 2)x}\{\sin nx + \sin (n + 2)x}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}} ight|$.Using $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$ and $\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$:$\cos nx + \cos (n + 2)x = 2\cos(n+1)x \cos x$ and $\sin nx + \sin (n + 2)x = 2\sin(n+1)x \cos x$.Subtracting $2\cos x$ times $C_2$ from $C_1$ and $C_3$ makes the determinant zero or simplifies it. Alternatively,calculating the determinant directly:$\Delta = 1[\cos(n+1)x \sin(n+2)x - \sin(n+1)x \cos(n+2)x] - 1[\cos nx \sin(n+2)x - \sin nx \cos(n+2)x] + 1[\cos nx \sin(n+1)x - \sin nx \cos(n+1)x]$.Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$:$\Delta = \sin((n+2)x - (n+1)x) - \sin((n+2)x - nx) + \sin((n+1)x - nx)$.$\Delta = \sin(x) - \sin(2x) + \sin(x) = 2\sin x - \sin 2x$.Since the result $2\sin x - \sin 2x$ does not contain $n$,the determinant is independent of $n$.

The value of the determinant $\left| {\begin{array}{*{20}{c}}1&1&1\\{\cos (nx)}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin (nx)}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}} \right|$ is independent of:

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