Why is the Half-Angle formula derived from the Double-Angle formula?

Question

Why is the Half-Angle formula derived from the Double-Angle formula?

Isn't that like ending back to where you started? Half a double?

Obviously not or this wouldn't work! So can someone explain to a dummy like me? :)

Answer 1

cos(x) = cos(2*x/2) = 1 - sin^2(x/2) from the double-angle formula



then solving for the sin(x/2), cos(x) - 1 = sin^2(x/2) --> sin(x/2) = +- sqrt((cos(x) -1)/2)



similarly, cos(x) = cos(2*x/2) = 2cos^2(x/2) - 1



solve for cos(x/2) --> cos(x) + 1 = 2cos^2(x/2) --> (cos(x) + 1)/2 = cos^2(x/2)

Answer 2

My question to your "why?" is "why not." Yes, you could derive them geometrically just how the double angled ones were derived, but the double angle identity came first, and it's an identity, thus true for every possible input. This happens quite a bit. If you want to see the geometric derivation here's a link. I didn't read through it because it was too long (hence why we use the identities).

http://feucht.us/writings/halfdoubleangle.php

Answer 3

Let the equal sides of the triangle be of size a let the attitude between these two facets be x drop a perpendicular from the vertex to the bottom. Now u have 2 equivalent triangles the attitude between the perpendicular and probably the most equal aspects is now (x/2) length of perpendicular = a cos(x/2) size of base = 2a sin(x/2) area of triangle = (half) 2sin(x/2) cos(x/2) a-square = (1/2) (sin x) a-square that is it!

Answer 4

https://answersrip.com/question/index?qid=20121029073434AADRyrZ