3 ms·
You can derive the identity 2 sin^2(θ / 2) = 1 - cos(θ) mentioned in the article from the double angle cosine formula, cos(2 Φ) = cos^2(Φ) - sin^2(Φ) Sinc
by picomancer 13y ago
You can derive the identity 2 sin^2(θ / 2) = 1 - cos(θ) mentioned in the article from the double angle cosine formula,
cos(2 Φ) = cos^2(Φ) - sin^2(Φ)
Since cos^2(Φ) + sin^2(Φ) = 1, substitute 1 - sin^2(Φ) for cos^2(Φ) in the above and you have:
cos(2 Φ) = cos^2(Φ) - sin^2(Φ)
= (1 - sin^2(Φ)) - sin^2(Φ)
= 1 - 2 sin^2(Φ)
2 sin^2(Φ) = 1 - cos(2 Φ)
Let θ = 2 Φ and the above becomes:
2 sin^2(θ / 2) = 1 - cos(θ)
- picomancer 13y agoIf you don't remember the double angle formula I used in the parent, but you remember the angle-addition formulas: cos(ψ + Φ) = cos(ψ) cos(Φ) - sin(ψ) sin(Φ) sin(ψ + Φ) = sin(ψ) cos(Φ) + cos(ψ) sin(Φ) You can just set ψ = Φ and get: cos(2 Φ) = cos^2(Φ) - sin^2(Φ) sin(2 Φ) = 2 cos(Φ) sin(Φ) If you don't remember the angle-addition formulas, but you remember that multiplying two complex numbers means adding their angles (arguments) and multiplying their lengths (moduli), you can just pick two unit-modulus complex numbers w = cos(ψ) + i sin(ψ) = a + bi, and z = cos(Φ) + i sin(Φ) = c + di. Multiplying them gives some complex number wz = u + vi, but because these are unit vectors, it must be the case that u = cos(ψ + Φ) and v = sin(ψ + Φ). So u + vi = wz = (a + bi) (c + di) = ac + (bc + ad)i + bdi^2 = (ac - bd) + (bc + ad)i Then equating real and imaginary parts, you get u = ac - bd v = bc + ad If you substitute the trig expressions for the variables it becomes cos(ψ + Φ) = cos(ψ) cos(Φ) - sin(ψ) sin(Φ) sin(ψ + Φ) = sin(ψ) cos(Φ) + cos(ψ) sin(Φ) Math is good for people who can't remember things, because you can always re-derive anything you've forgotten.