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Find the coordinate matrix of the function $\cos^2x$ relative to the ordered basis $\{1,\cos x,\sin x,\sin 2x\}$.

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    Should that $\sin 2x$ maybe be $\sin^2 x$?2012-12-15

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Assume $\tag1\cos^2x=a+b\cos x+c\sin x+d\sin 2x.$ Then $\tag 2\cos 2x = 2\cos^2x-1=(2a-1)+2b\cos x+2c\sin x+2d\sin 2x.$ Taking derivatives of $(2)$, we obtain $\tag3-2\sin 2x =-2b\sin x+2c\cos x+4d\cos 2x.$ Solving $(3)$ for $2d\cos2x$, we find $\tag42d\cos 2x=b\sin x-c\cos x-\sin 2x$ and by equating $(4)$ and $2d(2)$ $\tag5 0=2d(2a-1)+(4d-1)b\cos x+(4d+1)c\sin x+(4d^2+1)\sin2x.$

Taking the linear independece of $\{1, \cos x,\sin x,\sin 2x\}$ for granted (which is correct), we read $4d^2+1=0$ from $(5)$, which is not possible in the reals. But even allowing complex coefficients, we also conclude $b=c=0$ and $a=\frac12$, which contradicts what we obtain from letting $x=0$ in $(1)$, namely $1=a+b$.