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By using $(\cos x+i\sin x)^3=\cos 3x+ i \sin 3x$, We have:

$\sin ^3x= \frac{3}{4} \sin x- \frac{1}{4} \sin 3x$

$\cos ^3 x=\frac{3}{4}\cos x + \frac{1}{4}\cos 3x$

If we compare the coefficients of the equations we see coefficient of $\sin x$ and $\cos x$ are both $\frac{3}{4}$ and coefficient of $\sin 3x $ and $\cos 3x$ are $-\frac{1}{4}$ and $\frac{1}{4}$ respectively. the coefficients are very similar to each other !

Is there any approach to make me understand by intuition that why the coefficients are so similar to each other in these two equations?

bjcolby15
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Aligator
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    Are you sure your calculations are correct? As far as I know $\sin 3x$ cannot be expressed as a polynomial of $\sin x$. – Trebor Sep 05 '20 at 17:09
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    @Trebor You've formulated that correctly: "As far as I know". –  Sep 05 '20 at 17:12
  • @Trebor Yes see this: https://math.stackexchange.com/questions/97654/is-sin3-x-frac34-sin-x-frac14-sin-3x – Aligator Sep 05 '20 at 17:13
  • @Trebor $$\sin{2x}=2\sin{x}\cos{x}\implies \sin{3x}=\sin(2x+x)=\sin2x\cos{x}+\sin{x}\cos{2x}=2\sin{x}\cos^2{x}+\sin{x}(1-2\sin^2{x})=2\sin{x}(1-\sin^2{x})+\sin{x}(1-2\sin^2{x})=3\sin{x}-4\sin^3{x}$$ – A-Level Student Sep 06 '20 at 11:26

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Simply do the substitution $x \mapsto x + \frac \pi2$: Then $\sin x$ becomes $\sin (x + \frac\pi2)$, which is $\cos x$. And $\sin 3x$ becomes $\sin (3x + \frac{3\pi}2)$, which is $-\cos 3x$.

Note that similar substitutions do not exist for $\sin 2x$ and $\cos 2x$, so there's nothing similar for $\sin 2x$. That's where I got the false impression that you can't express $\sin 3x$ as a polynomial of $\sin x$.

Trebor
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  • Yep. The substitution notation $x \mapsto y$ is actually a bit ambiguous, so I got it the wrong way round. – Trebor Sep 05 '20 at 17:19
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    There's hope this will be ok after a few more edits. –  Sep 05 '20 at 17:20