Using $$\langle \hat S_x \rangle ={\chi_{s}}^{\dagger}(t) \hat S_x \chi_s(t)=\frac{\hbar}{2}{\chi_{s}}^{\dagger}(t)\sigma_x\chi_s(t)\tag{1}$$ where the time dependent $2$ - component spin vector is $$\chi_s(t)=\begin{bmatrix}\chi_1(t) \\ \chi_2(t)\end{bmatrix}$$ and $$\chi_1(t)=a_1\exp\left(\frac{-i\mu_BB\,t}{\hbar}\right)$$ $$\chi_2(t)=a_2\exp\left(\frac{i\mu_BB\,t}{\hbar}\right)$$ and the Pauli spin matrix $$\sigma_x=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$$
$\mu_B$ is the Bohr magneton.
So from $(1)$
$$\begin{align}\langle \hat S_x \rangle &=\frac{\hbar}{2}{\chi_{s}}^{\dagger}(t)\sigma_x\chi_s(t)\\&= \frac{\hbar}{2}\begin{pmatrix}{\chi_1}^*(t) & {\chi_2}^*(t)\end{pmatrix}\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}\begin{pmatrix}{\chi_1}(t) \\ {\chi_2}(t)\end{pmatrix}\\&=\frac{\hbar}{2}\begin{pmatrix}{\chi_1}^*(t) & {\chi_2}^*(t)\end{pmatrix}\begin{pmatrix}{\chi_2}(t) \\ {\chi_1}(t)\end{pmatrix}\\&=\frac{\hbar}{2}\left({\chi_1}^*(t){\chi_2}(t)+{\chi_2}^*(t){\chi_1}(t)\right)\\&=\frac{\hbar}{2}\left({a_1}^*\exp\left(\frac{i \mu_BB\, t}{\hbar}\right)a_2\exp\left(\frac{i \mu_BB\, t}{\hbar}\right)+{a_2}^*\exp\left(\frac{-i \mu_BB\, t}{\hbar}\right)a_1\exp\left(\frac{-i \mu_BB\, t}{\hbar}\right)\right)\\&=\frac{\hbar}{2}\left({a_1}^*a_2\exp\left(\frac{2i \mu_BB\, t}{\hbar}\right)+a_1{a_2}^*\exp\left(\frac{-2i \mu_BB\, t}{\hbar}\right)\right)\end{align}$$
This is as far as I can seem to get. I am trying to make use of the result that $$2\cos\left(\frac{2\mu_BB\, t}{\hbar}\right)=\exp\left(\frac{2i \mu_BB\, t}{\hbar}\right)+\exp\left(\frac{-2i \mu_BB\, t}{\hbar}\right)$$ I note that the second term in the bracket of the last equation of $\langle \hat S_x \rangle$ is the complex conjugate of the first term in that bracket. Even with this knowledge, I am still confused about how to complete this proof.
Could someone please provide me with any hints or advice to prove that $$\langle \hat S_x \rangle=\hbar\,a_1a_2\cos\left(\frac{2\mu_BB\, t}{\hbar}\right)?$$