3-tap FIR filter: simple expression for $H(e^{j\omega})$ using trigonometric identities
HINT:

Write $H(e^{j\omega})=e^{-j\omega}G(e^{j\omega})$ and use $(e^{j\omega}+e^{-j\omega})=2\cos(\omega)$. $G(e^{j\omega})$ should turn out to be real-valued.