Exercise:

A star’s true position is
Right Ascension 6h 0m 0s, declination 0° 0' 0",
and it lies at a distance of 25 parsecs.
On the date of the Spring Equinox,
how far will it appear to be shifted by annual parallax,
and in what direction?

First convert from RA and dec. into ecliptic coordinates.
sin(δ) = sin(β) cos(ε) + cos(β) sin(ε) sin(λ)
sin(β) = sin(δ) cos(ε) - cos(δ) sin(ε) sin(α)
cos(λ) cos(β) = cos(α) cos(δ)
where α = 6h0m = 90°, δ = 0°

So this star has λ = 90°, β = -ε = -23°26'.

Its distance = 25 pc, so annual parallax Π = (1/25)" = 0.040"

At the spring equinox, the Sun has ecliptic longitude λ_S = 0°
so λ_S-λ = -90°.

Δλcosβ = Π sin(λ_S-λ) = 0.040"
cosβ = 0.918
so Δλ = 0.044".
Δβ= -Π cos(λ_S-λ) sinβ = 0

So star is shifted 0.044" eastwards, by parallax.

Back to "annual parallax".