Edit
Following the comment of @r9m below, if we differentiate equation $(9)$ below w.r. to $x$ we obtain
$$ \begin{aligned} &\frac{d}{dx}\left(\frac{1}{2^{2n}}\mathrm{Cl}_{2n+1}(2 x)=\mathrm{Cl}_{2n+1}( x)+\mathrm{Cl}_{2n+1}\left( \pi-x\right)\right)\\ &\frac{1}{2^{2n}} \sum_{k=1}^\infty \frac{-2k\sin(2kx)}{k^{2n+1}}=\sum_{k=1}^\infty \frac{-k\sin(kx)}{k^{2n+1}}+\sum_{k=1}^\infty \frac{-k\sin(k(\pi-x))}{k^{2n+1}}\\ &\mathrm{Cl}_{2n}(2 x)=2^{2n-1}\left(\mathrm{Cl}_{2n}( x)-\mathrm{Cl}_{2n}\left( \pi-x\right)\right) \qquad \blacksquare \end{aligned} $$
The Clausen function is defined as
$$ \mathrm{Cl}_{n}(x) \equiv \begin{cases}S_{n}(x)=\sum_{k=1}^{\infty} \frac{\sin (k x)}{k^{n}} & n \text { even } \\ C_{n}(x)=\sum_{k=1}^{\infty} \frac{\cos (k x)}{k^{n}} & n \text { odd },\end{cases} $$
Recall the Fourier expansion
$$\ln\left(2 \sin\left( \frac{x}{2}\right) \right)=-\sum_{n=1}^\infty \frac{\cos\left( n x\right)}{n} \tag{1}$$
Integrating both sides of $(1)$ from $0$ to $x$ we obtain
$$\begin{align} \int_0^x\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds&=-\sum_{n=1}^\infty \frac{\sin\left( n x\right)}{n^2}\\ &=-\operatorname{Cl}_2(x) \tag{2} \end{align} $$
From this representation $(2)$ we can derive a duplication formula
$$\operatorname{Cl}_2(2x)=2\operatorname{Cl}_2(x)-2\operatorname{Cl}_2(\pi-x) \tag{3}$$
Proof:
$$ \begin{aligned} \operatorname{Cl}_2(2x)&=-\int_0^{2x}\ln\left(2 \sin\left( \frac{t}{2}\right) \right)\,dt\\ &=-2\int_0^{x}\ln\left(2 \sin\left( t\right) \right)\,dt\\ &=-2\int_0^{x}\ln\left(4 \sin\left( \frac{t}{2}\right)\cos\left( \frac{t}{2}\right) \right)\,dt\\ &=-2\int_0^x\ln\left(2 \sin\left( \frac{t}{2}\right) \right)\,dt-2\int_0^x\ln\left(2 \cos\left( \frac{t}{2}\right) \right)\,dt\\ &=2\operatorname{Cl}_2(x)-4\int_0^{x/2}\ln\left(2 \cos\left(t\right) \right)\,dt\\ &=2\operatorname{Cl}_2(x)-4\int_{\frac{\pi-x}{2}}^{\pi/2}\ln\left(2 \sin\left(t\right) \right)\,dt\\ &=2\operatorname{Cl}_2(x)-4\int_{0}^{\pi/2}\ln\left(2 \sin\left(t\right) \right)\,dt+4\int_{0}^{\frac{\pi-x}{2}}\ln\left(2 \sin\left(t\right) \right)\,dt\\ &=2\operatorname{Cl}_2(x)-2\int_{0}^{\pi}\ln\left(2 \sin\left(\frac{t}{2}\right) \right)\,dt+2\int_{0}^{\pi-x}\ln\left(2 \sin\left(\frac{t}{2}\right) \right)\,dt\\ &=2\operatorname{Cl}_2(x)+2\operatorname{Cl}_2(\pi)-2\operatorname{Cl}_2(\pi-x)\\ &=2\operatorname{Cl}_2(x)-2\operatorname{Cl}_2(\pi-x) \qquad \blacksquare\\ \end{aligned} $$
Following the same procedure we can obtain a duplication formula for $\operatorname{Cl}_3(x)$
$$\operatorname{Cl}_3(2x)=4\operatorname{Cl}_3(x)+4\operatorname{Cl}_3(\pi-x) \tag{4}$$
Proof:
Integrating both sides of $(3)$ form $0$ to $x$ we have:
$$\int_0^x\operatorname{Cl}_2(2t)\,dt=2\int_0^x\operatorname{Cl}_2(t)\,dt-2\int_0^x\operatorname{Cl}_2(\pi-t)\,dt$$
Lets evaluate each of these integrals separately
$$ \begin{aligned} \int_0^x \operatorname{Cl}_2(2t)\,dt&=\frac12\int_{0}^{2x} \operatorname{Cl}_2(t)\,dt\\ &=\frac12\left(\zeta(3)- \operatorname{Cl}_3(2x)\right) \qquad \blacksquare \end{aligned} $$
$$ \begin{aligned} \int_0^x \operatorname{Cl}_2(t)\,dt&=\zeta(3)-\operatorname{Cl}_3(x) \qquad \blacksquare \end{aligned} $$
$$ \begin{aligned} \int_0^x \operatorname{Cl}_2(\pi-t)\,dt&=\int_{\pi-x}^{\pi} \operatorname{Cl}_2(t)\,dt\\ &=\int_0^\pi \operatorname{Cl}_2(t)\,dt-\int_0^{\pi-x} \operatorname{Cl}_2(t)\,dt\\ &=\zeta(3)-\operatorname{Cl}_3(\pi)-\left(\zeta(3)-\operatorname{Cl}_3(\pi-x) \right)\\ &=-\sum_{n=1}^\infty\frac{\cos(n\pi)}{n^3}+\operatorname{Cl}_3(\pi-x) \\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^3}+\operatorname{Cl}_3(\pi-x) \\ &=\eta(3)+\operatorname{Cl}_3(\pi-x) \\ &=\frac34\zeta(3)+\operatorname{Cl}_3(\pi-x) \qquad \blacksquare\\ \end{aligned} $$
Putting all together
$$ \begin{aligned} &\frac12\left(\zeta(3)- \operatorname{Cl}_3(2x)\right)=2\left( \zeta(3)-\operatorname{Cl}_3(x)\right)-2\left( \frac34\zeta(3)+\operatorname{Cl}_3(\pi-x) \right)\\ &\operatorname{Cl}_3(2x)=4\operatorname{Cl}_3(x)+4\operatorname{Cl}_3(\pi-x) \qquad \blacksquare \end{aligned} $$
According to Wikipedia we have a general formula for the duplication formula dor even and odd indices, namely:
$$ \mathrm{Cl}_{m+1}(2 \theta)=2^{m}\left[\mathrm{Cl}_{m+1}(\theta)+(-1)^{m} \mathrm{Cl}_{m+1}(\pi-\theta)\right] \tag{5} $$
For odd indices I was able to establish the general formula as follows
We have (very easy to prove just by expanding the R.H.S.)
$$2\sum_{k=1}^\infty a_{2k}=\sum_{k=1}^\infty a_{k}+\sum_{k=1}^\infty (-1)^ka_{k} \tag{6}$$
Then
$$ \begin{align} 2\sum_{k=1}^\infty \frac{\cos\left(2x k \right)}{(2k)^{2n+1}}&=\frac{1}{2^{2n}}\sum_{k=1}^\infty \frac{\cos\left(2x k \right)}{k^{2n+1}}\\ &=\frac{1}{2^{2n}}\mathrm{Cl}_{2n+1}(2 x) \tag{7} \end{align} $$
On the other hand
$$ \begin{align} \sum_{k=1}^\infty \frac{\cos\left(x k \right)}{k^{2n+1}}+\sum_{k=1}^\infty \frac{\cos\left( k(\pi-x) \right)}{k^{2n+1}} &=\sum_{k=1}^\infty \frac{\cos\left(x k \right)}{k^{2n+1}}+\sum_{k=1}^\infty \frac{(-1)^k\cos\left( k x \right)}{k^{2n+1}}\\ &=\mathrm{Cl}_{2n+1}( x)+\mathrm{Cl}_{2n+1}\left( \pi-x\right) \tag{8} \end{align} $$
By $(6)$ we can equate $(7)$ and $(8)$ obtaining
$$\frac{1}{2^{2n}}\mathrm{Cl}_{2n+1}(2 x)=\mathrm{Cl}_{2n+1}( x)+\mathrm{Cl}_{2n+1}\left( \pi-x\right) \tag{9}$$
And proving the duplication formula for odd indices.
For even indices I failed to prove using the same method because
$$ \begin{aligned} \sum_{k=1}^\infty \frac{\sin\left(x k \right)}{k^{2n}}+\sum_{k=1}^\infty \frac{\sin\left( k(\pi-x) \right)}{k^{2n}} &=\sum_{k=1}^\infty \frac{\sin\left(x k \right)}{k^{2n}}-\sum_{k=1}^\infty \frac{(-1)^k\sin\left( k x \right)}{k^{2n}}\\ &=\mathrm{Cl}_{2n+1}( x)+(-1)^{2n+1}\mathrm{Cl}_{2n+1}\left( \pi-x\right) \end{aligned} $$
But we can no longer use $(6)$ because
$$\sum_{k=1}^\infty a_{k}-\sum_{k=1}^\infty (-1)^ka_{k}=2\sum_{k=1}^\infty a_{2k-1}$$
Leading to
$$ 2\sum_{k=1}^\infty \frac{\cos\left((2 k-1)x \right)}{(2k-1)^{2n+1}} $$
For the L.H.S.
So, I would like to find a proof of the duplication formula for even indices .