Half-Range Fourier Cosine Series over Negative Range

Theorem
Let $\map f x$ be a real function defined on the open real interval $\openint 0 \lambda$.

Let $f$ be expressed using the half-range Fourier cosine series over $\openint 0 \lambda$:


 * $\displaystyle \map C x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos \frac {n \pi x} \lambda$

where:
 * $a_n = \displaystyle \frac 2 \lambda \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$

for all $n \in \Z_{\ge 0}$.

Then over the closed real interval $\openint {-\lambda} 0$, $\map C x$ takes the values:
 * $\map C x = \map f {-x}$

That is, the real function expressed by the half-range Fourier cosine series over $\openint 0 \lambda$ is an even function over $\openint {-\lambda} \lambda$.

Proof
From Fourier Series for Even Function over Symmetric Range, $\map C x$ is the Fourier series of an even real function over the interval $\openint 0 \lambda$.

We have that $\map C x \sim \map f x$ over $\openint 0 \lambda$.

Thus over $\openint {-\lambda} 0$ it follows that $\map C x = \map f {-x}$.