Question on othogonal projection on to a span
I am trying to answer the question above and I just want to confirm I am on the right track. I found an orthonormal basis $\{ \frac{4cosx}{\pi}, sinx - \frac{8cosx}{\pi^2} \}$ using the gram schmidt process. But the answer I got for the projection of $1$ onto $span(\{cosx, sinx\})$ using this basis is $\frac{16cosx}{\pi^2} + ((1 - \frac{4}{\pi^2}) \cdot sinx - \frac{8cosx}{\pi^2})$. However, I have seen another answer being $sinx+cosx$ but this is likely because they assumed $\{cosx, sinx\}$ was already an orthonormal basis. I checked however and the inner product of $<cosx, sinx>$ is not zero so I found this assumption weird to make. Am I wrong in the way I went about my answer? Am I missing something? Any help would be appreciated.
Easier, you want $a$ and $b$ so that $1-(a\cos x + b\sin x)$ is orthogonal to both $\cos x$ and $\sin x$. This gives $$1+(\pi/4)a+(1/2)b = 1+(1/2)a+(\pi/4)b=0.$$ Offhand, neither of the answers you gave seems reasonable.