**Appendix A**

**Proof of Lemma 1.** Putting *p* = 1 in Equation (2) and using the expansion of the characteristic function of *θ*, we ge<sup>t</sup> (henceforth, we denote *E*(*X*) and *V*(*X*) as expectation and variance of a random variable *X*, respectively, as usual)

$$E\cos\theta = \rho\cos\mu\_0\tag{A1}$$

$$\Rightarrow E(\mathcal{C}\_1) = E\left[\frac{1}{m}\sum\_{i=1}^m \cos\theta\_i\right] = \rho\cos\mu\_0$$

Again, putting *p* = 2 in Equation (2) and using the expansion of the characteristic function of *θ*, we ge<sup>t</sup>

$$E\cos 2\theta = \rho^{2^n} \cos 2\mu\_0 \tag{A2}$$

$$\implies E(\mathcal{C}\_2) = E\left[\frac{1}{m} \sum\_{i=1}^m \cos 2\theta\_i\right] = \rho^{2^n} \cos 2\mu\_0$$

> In addition, Equation (A2) implies that,

$$\begin{aligned} E\cos^2\theta &= \frac{(\rho^{2^a}\cos 2\mu\_0 + 1)}{2} \\ \Rightarrow V(\cos\theta) &= E\cos^2\theta - E^2\cos\theta \\ &= \frac{\rho^{2^a}\cos 2\mu\_0 + 1}{2} - \rho^2\cos^2\mu\_0 \end{aligned}$$

Hence,

$$V(\mathcal{C}\_1) = V\left[\frac{1}{m}\sum\_{i=1}^{m}\cos\theta\_i\right] \tag{A3}$$

$$=\frac{\rho^{2^{\ast}}\cos 2\mu\_0 + 1 - 2\rho^2 \cos^2 \mu\_0}{2m} \tag{A4}$$

Now, putting *p* = 4 in Equation (2) and using the expansion of the characteristic function of *θ*, we get,

$$E\cos 4\theta = \rho^{4^a} \cos 4\mu\_0 \tag{A5}$$

$$\implies E\left[\cos^2 2\theta\right] = \frac{\left(\rho^{4^a} \cos 4\mu\_0 + 1\right)}{2}$$

Hence,

$$\begin{split} V(\cos 2\theta) &= E \cos^2 2\theta - E^2 \cos 2\theta \\ &= \frac{\rho^{4^\ast} \cos 4\mu\_0 + 1}{2} - (\rho^{2^\ast})^2 \cos^2 2\mu\_0 \end{split}$$

Therefore,

$$\begin{split} V(\mathcal{C}\_2) &= V \left[ \frac{1}{m} \sum\_{i=1}^{m} \cos 2\theta\_i \right] \\ &= \frac{\rho^{4^u} \cos 4\mu\_0 + 1 - 2(\rho^{2^u})^2 \cos^2 2\mu\_0}{2m} \end{split} \tag{A6}$$

Now, putting *p* = 1 in Equation (2) and using the expansion of the characteristic function of *θ*, we ge<sup>t</sup>

$$E\sin\theta = \rho\sin\mu\_0\tag{A7}$$

$$\Rightarrow E(\mathcal{S}\_1) = E\left[\frac{1}{m}\sum\_{i=1}^m \sin\theta\_i\right] = \rho\sin\mu\_0$$

Again, putting *p* = 2 in Equation (2) and using the expansion of the characteristic function of *θ*, we ge<sup>t</sup>

$$E\sin 2\theta = \rho^{2^u} \sin 2\mu\_0 \tag{A8}$$

$$\implies E(\bar{S}\_2) = E\left[\frac{1}{m} \sum\_{i=1}^m \sin 2\theta\_i\right] = \rho^{2^u} \sin 2\mu\_0$$

> Now, using Equation (A2),

$$E\sin^2\theta = \frac{(1-\rho^{2^\ast}\cos2\mu\_0)}{2}$$

Hence,

$$\begin{split} V(\sin \theta) &= E \sin^2 \theta - E^2 \sin \theta \\ &= \frac{1 - \rho^{2^\ast} \cos 2\mu\_0}{2} - \rho^2 \sin^2 \mu\_0 \end{split}$$

Therefore,

$$\begin{aligned} V(\bar{S}\_1) &= V\left[\frac{1}{m} \sum\_{i=1}^m \sin \theta\_i\right] \\ &= \frac{1 - \rho^{2^x} \cos 2\mu\_0 - 2\rho^2 \sin^2 \mu\_0}{2m} \end{aligned}$$

Now, using Equation (A5),

$$E\sin^2 2\theta = \frac{(1 - \rho^{4^a}\cos 4\mu\_0)}{2}$$

Hence,

$$\begin{split} V(\sin 2\theta) &= E \sin^2 2\theta - E^2 \sin 2\theta \\ &= \frac{1 - \rho^{4^\kappa} \cos 4\mu\_0}{2} - (\rho^{2^\kappa})^2 \sin^2 2\mu\_0 \end{split}$$

Therefore,

$$\begin{aligned} V(\mathcal{S}\_2) &= V \left[ \frac{1}{m} \sum\_{i=1}^m \sin 2\theta\_i \right] \\ &= \frac{1 - \rho^{4^x} \cos 4\mu\_0 - 2(\rho^{2^x})^2 \sin^2 2\mu\_0}{2m} \end{aligned}$$

Now, using Equations (A1), (A7) and (A8),

$$\begin{aligned} Cov(\cos \theta, \sin \theta) &= E \cos \theta \sin \theta - E \cos \theta E \sin \theta \\ &= \frac{\rho^{2^n} \sin 2\mu\_0}{2} - \rho \cos \mu\_0 \rho \sin \mu\_0 \end{aligned}$$

Therefore,

$$E\sum\_{i=1}^{m}\cos\theta\_{i}\sum\_{i=1}^{m}\sin\theta\_{i} = E\sum\_{i=1}^{m}\cos\theta\_{i}\sin\theta\_{i} + \sum\_{i}^{m}\sum\_{j\neq i}^{m}\cos\theta\_{i}\sin\theta\_{j}\delta\_{j}$$

> Thus,

$$\begin{aligned} Cov(\vec{C}\_1, \vec{S}\_1) &= Cov\left[\frac{1}{m} \sum\_{i=1}^m \cos \theta\_{i\prime} \frac{1}{m} \sum\_{i=1}^m \sin 2\theta\_i \right] \\ &= \frac{\rho^{2^\ast} \sin 2\mu\_0 - 2\rho^2 \cos \mu\_0 \sin \mu\_0}{2m} \end{aligned}$$

Putting *p* = 3 in Equation (2) and using the expansion of characteristic function of *θ*, we ge<sup>t</sup>

$$E\left(\sin 3\theta\right) = \rho^{\mathcal{I}^x} \sin 3\mu\_0 \tag{A9}$$

Now,

$$\begin{aligned} \text{Cov}(\vec{C}\_1, \vec{S}\_2) &= \text{Cov}\left[\frac{1}{m} \sum\_{i=1}^m \cos \theta\_{i\prime} \frac{1}{m} \sum\_{i=1}^m \sin 2\theta\_i\right] \\ \text{Cov}(\cos \theta, \sin 2\theta) &= E \cos \theta \sin 2\theta - E \cos \theta E \sin 2\theta \end{aligned}$$

Now, using Equations (A7) and (A9),

$$\begin{aligned} E \cos \theta \sin 2\theta &= E \left[ \frac{\sin 3\theta + \sin \theta}{2} \right] \\ &= \frac{\rho^{3^\ast} \sin 3\mu\_0 + \rho \sin \mu\_0}{2} \end{aligned}$$

Thus, using Equations (A1) and (A8),

$$Cov(\cos \theta, \sin 2\theta) = \frac{\rho^{3^u} \sin 3\mu\_0 + \rho \sin \mu\_0}{2} - \rho \cos \mu\_0 \rho^{2^u} \sin 2\mu\_0$$

Hence,

$$Cov(\mathcal{C}\_1, \mathcal{S}\_2) = \frac{\rho^{3^\ast} \sin 3\mu\_0 + \rho \sin \mu\_0 - 2\rho^{2^\ast + 1} \cos \mu\_0 \sin 2\mu\_0}{2m}.$$

Similarly, it can be shown that,

$$\text{Cov}(\mathcal{C}\_1, \mathcal{C}\_2) = \frac{\rho \cos \mu\_0 + \rho^{3^\ast} \cos 3\mu\_0 - 2\rho^{2^\ast + 1} \cos \mu\_0 \cos 2\mu\_0}{2m} \tag{A10}$$

$$\text{Cov}(\mathcal{C}\_2, \mathcal{S}\_1) = \frac{\rho^{\otimes^\alpha} \sin \mathfrak{z} \mu\_0 - \rho \sin \mu\_0 - 2\rho^{2^\ast + 1} \cos 2\mu\_0 \sin \mu\_0}{2m}$$

$$\text{Cov}(\mathcal{C}\_2, \mathcal{S}\_2) = \frac{\rho^{4^\ast} \sin 4\mu\_0 - 2(\rho^{2^\ast})^2 \cos 2\mu\_0 \sin 2\mu\_0}{2m}$$

$$Cov(\bar{S}\_1, \bar{S}\_2) = \frac{\rho \cos \mu\_0 - \rho^{\mathfrak{J}^d} \cos 3\mu\_0 - 2\rho^{2^d + 1} \sin \mu\_0 \sin 2\mu\_0}{2m}$$
