Why is (cosθ1.cosθ2 - sinθ1.sinθ2) equal to cos(θ1+θ2)?

Question

Why is (cosθ1.cosθ2 - sinθ1.sinθ2) equal to cos(θ1+θ2)?

And why is (sinθ1.cosθ2 + cosθ1.sinθ2) equal to sin(θ1+θ2)?

Note: I've seen this in a chapter about complex numbers.

Answer 1

If you accept De Moivre's theorem as a basis for this, it's pretty easy: take two complex numbers with modulus 1 and arguments θ1 and θ2. In rectangular coordinates these are
cos θ1 + i sin θ1 and cos θ2 + i sin θ2. When you multiply them DM's theorem says you get a complex number with modulus 1 and argument θ1+θ2, so we have
cos(θ1+θ2) + i sin (θ1+θ2) = (cos θ1 + i sin θ1) (cos θ2 + i sin θ2)
= (cos θ1 cos θ2 - sin θ1 sin θ2) + i (sin θ1 cos θ2 + cos θ1 sin θ2)
so by equating real parts we get
cos (θ1+θ2) = cos θ1 cos θ2 - sin θ1 sin θ2
and by equating imaginary parts we get
sin (θ1+θ2) = sin θ1 cos θ2 + cos θ1 sin θ2.

If you're feeling adventurous you can derive this directly from the definition of sin and cos as complex exponentials:
cos x = 1/2 (e^(ix) + e^(-ix))
sin x = 1/(2i) (e^(ix) - e^(-ix))

so e.g. cos x cos y - sin x sin y
= (1/4) [(e^(ix) + e^(-ix)) . (e^(iy) + e^(-iy))] - (1/4i^2) [(e^(ix) - e^(-ix)) . (e^(iy) - e^(-iy))]
= (1/4) [e^(ix+iy) + e^(ix-iy) + e^(-ix+iy) + e^(-ix-iy)] + (1/4) [e^(ix+iy) - e^(ix-iy) - e^(-ix+iy) + e^(-ix+iy)]
= (1/4) [2e^(ix+iy) + 2e^(-ix-iy)]
= (1/2) [e^(i(x+y)) + e^(-i(x+y))]
= cos (x+y).

There is also a geometric proof, but that doesn't have anything to do with complex numbers.

Note to the previous answerers: trig identities also have to be proven. They don't just appear out of nowhere as gospel truth.

Answer 2

i'm the boss of all my equines. we are no longer equivalent; there is by no skill a difficulty wherein they get to make certain what we are able to do. If I say we are going for a trip, we are going for a trip. If I say i choose for to hold that foot off the floor, that foot is coming up. that's no longer a talk, that's no longer an equivalent partnership wherein we weigh up the end results of a handful of ideas: what I say is going. that doesn't recommend i do no longer cope with my horses and ponies with admire. i think we've a mutual admire and it is going to continuously be a 2 way communication in that I hear to what's being suggested; i'm no longer dominant in a forceful way, regardless of if it somewhat is critical to a working dating and a sensible, chuffed horse that i'm respected and accompanied by using fact the boss. the secret's they are no longer people and we can't have conversations approximately it. they're horses - working animals, no longer ultimate pals - and correct coaching is significant.

Answer 3

It is an a trig identify just like sin²x + cos²x = 1

Answer 4

its double angle

thats just a property of sin and cos

plugin any values for theta 1 and theta 2 and you'll see it always works