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Can we write $\cos^2(x)$ as linear combination of $x \mapsto \sin(x)$ and $x \mapsto \cos(x)$?

I know $$ \cos^2(x) = \frac{\cos(2x) + 1}{2} = 1 - \sin^2(x) = \cos(2x) + \sin^2(x) $$ but none of these helped. Then, I tried to solve $$ \cos^2(x) = \alpha \sin(x) + \beta \cos(x) $$ for the coefficients $\alpha, \beta \in \mathbb{R}$. But when plugging in $x = 0$ I get $\beta = 1$ and for $x = \frac{\pi}{2}$ I get $\alpha = 0$. Plugging those values back in I obtain a false statement, and WolframAlpha can't do better!

This is from a numerical analysis exam and the second function is $x \mapsto \sqrt{2}\cos\left(\frac{\pi}{4} - x \right)$, which can easily be expressed in terms of $x \mapsto \sin(x)$ and $x \mapsto \cos(x)$ by the corresponding addition formula.

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    $\begingroup$ Using the argument with $\alpha,\beta$ you have correctly shown that this is impossible. $\endgroup$ – Wojowu Feb 26 at 19:56
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Say we can do that, then

$$\cos^2x-b\cos x = a\sin x$$ Let $t= \cos x$ and square this equation. We get $$ t^4-2bt^3+b^2t^2 = a^2-a^2t^2$$ which should be valid for all $t\in[-1,1]$ and there for for all $t$. So the polynomials must be equal for all $t$, so by comparing the coeficients we get $1=0$. A contradiction.

So you can not express $\cos ^2x$ as linear combination of $\cos x$ and $\sin x$.

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The function $f(x):=\cos^2 x$ has $f(x+\pi)\equiv f(x)$, but any linear combination $g$ of $\cos$ and $\sin$ has $g(x+\pi)\equiv -g(x)$.

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You just stopped too early.

Assume $\cos^2x=a\sin x+b\cos x$.

  • Evaluating at $0$ yields $b=1$.
  • Evaluating at $\pi$ yields $b=-1$.

End.

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