Consider a star moving with velocity $v$ at an angle $\theta$ with respect to its line of sight to Earth. Show that the relativistic Doppler shift is
$$\lambda_{obs} = \frac{1 - \frac{v}{c} cos(\theta)}{\sqrt{1 - \frac{v^2}{c^2}}} \lambda_{em}$$
in which $c$ is the speed of light, $\lambda_{obs}$ is the observed wavelength, and $\lambda_{em}$ is the emitted wavelength.
Can someone show me to derive this equation? So far, I have been using a reference frame $S'$ for a certain angle $\theta'$ in which the $y' = ct'\sin(\theta') $ and $x' = ct'\cos(\theta')$. I used the Lorentz transformation to find that
$$x = \frac{x' + vt'}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{ct'\cos(\theta')}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{ct'(\cos(\theta') + \frac{v}{c})}{\sqrt{1 - \frac{v^2}{c^2}}} $$
I am not sure what do from here. Also, what happens for velocities that are much smaller than c? How can I use this equation to write how at $v\ll c$ the equation reduces to the usual expression for a Doppler shift such that
$$\lambda_{obs} = (1 + \frac{v_r}{c})\lambda_{em}$$
in which $v_r$ is the radial velocity?
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