<s>[INST] If $a_1+a_2\sin x+a_3\cos x+a_4\sin 2x+a_5\cos 2x=0$ is an identity in $x$,then prove that $(a_1,a_2,a_3,a_4,a_5)=(0,0,0,0,0)$ If $a_1+a_2\sin x+a_3\cos x+a_4\sin 2x+a_5\cos 2x=0$ is an identity in $x$,then prove that $(a_1,a_2,a_3,a_4,a_5)=(0,0,0,0,0)$
I tried:$a_1+a_2\sin x+a_3\cos x+a_4\sin 2x+a_5\cos 2x=0$
$a_1+a_2\sin x+a_3\cos x+2a_4\sin x\cos x+a_5(2\cos^2x-1)=0$
$a_1-a_5+a_2\sin x+a_3\cos x+2a_4\sin x\cos x+2a_5\cos^2x=0$
How do i prove that $a_1,a_2,a_3,a_4,a_5$ are all zero?
 [/INST] By repeated differentiating we obtain
$$\tag1 a_2\cos x-a_3\sin x+2a_4\cos 2x-2a_5\sin 2x=0$$
$$\tag2 -a_2\sin x-a_3\cos x-4a_4\sin 2x-4a_5\cos 2x=0$$
$$\tag3 -a_2\cos x+a_3\sin x-8a_4\cos 2x+8a_5\sin 2x=0$$
So by adding $(1)+(3)$
$$ 6a_5\sin 2x-6a_4\cos2x=0.$$
At $x=0$ this gives us $a_4=0$, and also $a_5=0$ from $x=\frac\pi4$.
With this, $(1)$ simplifies to
$$ a_2\cos x-a_3\sin x=0$$
whence we find $a_2=0$ from $x=0$ and $a_3=0$ from $x=\frac\pi2$.
Plug into the original equation to obtain $a_1=0$.
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