# Write $\cos^2(x)$ as linear combination of $x \mapsto \sin(x)$ and $x \mapsto \cos(x)$

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.

• Using the argument with $\alpha,\beta$ you have correctly shown that this is impossible. – Wojowu Feb 26 at 19:56

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 $$\sin x$$.

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

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.