<Dl> <Dd> cos ⁡ θ ≈ 1 − θ 2 2 (\ displaystyle \ cos (\ theta) \ approx 1 - (\ frac (\ theta ^ (2)) (2))) </Dd> </Dl> <Dd> cos ⁡ θ ≈ 1 − θ 2 2 (\ displaystyle \ cos (\ theta) \ approx 1 - (\ frac (\ theta ^ (2)) (2))) </Dd> <P> The opposite leg, O, is approximately equal to the length of the blue arc, s . Gathering facts from geometry, s = Aθ, from trigonometry, sin θ = O / H and tan θ = O / A, and from the picture, O ≈ s and H ≈ A leads to: </P> <Dl> <Dd> sin ⁡ θ = O H ≈ O A = tan ⁡ θ = O A ≈ s A = A θ A = θ . (\ displaystyle \ sin \ theta = (\ frac (O) (H)) \ approx (\ frac (O) (A)) = \ tan \ theta = (\ frac (O) (A)) \ approx (\ frac (s) (A)) = (\ frac (A \ theta) (A)) = \ theta .) </Dd> </Dl>

For which value of theta is tan theta equal to sin theta