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If 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(
by picomancer 13y ago
If 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.