Killing vectors - Schwarzschild metric Given the Schwarzschild metric, $$\mathrm{d}s^{2}=-\left(1-\frac{R_s}{r}\right)\mathrm{d}t^{2}+\left(1-\frac{R_s}{r}\right)^{-1}\mathrm{d}r^{2}+r^{2}\mathrm{d}\theta^{2}+ r^2 \sin^{2}\theta\mathrm{d}\phi^{2},$$ I'm asked to show that $$K^{\mu}=\left(1,0,0,0\right),\; R^{\mu}=(0,0,0,1)$$ are Killing vectors, i.e. they satisfy the Killing equation $$\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu=0.$$

First of all, using the metric I lower the indices: $$K_\mu=\left(-\left(1-\frac{R_s}{r}\right),0,0,0\right),\; R_\mu=\left(0,0,0,r^2\sin^2\theta\right).$$ Then, for $R_\mu$ I tried to compute $$\nabla_\mu R_\nu\equiv\partial_\mu R_\nu - \Gamma^\lambda_{\mu\nu} R_\lambda=\partial_\mu R_\nu - \Gamma^\phi_{\mu\nu} R_\phi;$$ so $$\nabla_r R_\phi=r\sin^2\theta, \; \nabla_\theta R_\phi=r^2\sin\theta\cos\theta.$$ But now it's not clear for me what should I do next.
 A: Your equation for $R_\mu$ reads $$\nabla_\mu R_\nu + \nabla_\nu R_\mu =\left(\partial_\mu R_\nu -\Gamma^\lambda_{\mu\nu} R_\lambda\right) + \left(\partial_\nu R_\mu - \Gamma^\sigma_{\nu\mu}R_\sigma\right)=\partial_\mu R_\nu + \partial_\nu R_\mu -2\Gamma^\phi_{\mu\nu} R_\phi.$$
If you use $\mu=r$ and $\mu=\theta$ (the only non-vanishing terms), you will find that Killing equation for $R_\mu$ it's satisfied.
A: $\def\bg{{\mathbf g}}$
It depends on how Killing vector has been defined. If definition was
"a vector field obeying Killing equation", then you are only left to
pursue the computation and I can add nothing. But there is an
alternative definition, which I prefer:
A Killing field is one whose flow is an isometry for all values of
parameter.
This requires to know what is the flow of a vector field and what
means "isometry" in a Riemannian manifold. I shortly summarize.
The flow of a vector field $X$ is simply the ensemble of its integral
curves, each parametrized by a real variable which I will name $u$.
For each value of $u$ the flow of $X$ defines a mapping $\mu$ of the
manifold into itself.
An isometry of a Riemannian manifold is a (differentiable) mapping
leaving invariant the metric tensor: $\mu^*\bg = \bg$.
Then it can be shown that $X$ is a Killing field for the metric $\bg$
iff it satisfies the Killing equation (where covariant derivative is
defined by Levi-Civita connection of $\bg$).
In your problem the flow of $K$ is obviously the mapping
$$\mu_u: (t,r,\theta,\phi) \mapsto (t+u,r,\theta,\phi)$$
i.e. a time translation by $u$. Since $\bg$ (i.e. $ds^2$) doesn't
depend on $t$, $\mu_u$ is an isometry for all $u$.
Analogously, the flow of $R$ is 
$$\nu_u: (t,r,\theta,\phi) \mapsto (t,r,\theta,\phi+u)$$
i.e. a rotation of angle $u$. And $\bg$ doesn't even depend on $\phi$, so $\nu_u$ is an isometry as well.
