Working on yet another physics exercise (more on that later, I hope), I found myself needing the trigonometric formulas for half angles.
Demonstration below the fold....
Aside: long ago, I demonstrated some of the consequences of the Law of Cosines.
One of those consequences is the double angle formula for cosines
cos( 2x )
= cos ( x - (-x) )
= cos(x)cos(-x) + sin(x)sin(-x)
= cos(x)^2 - sin(x)^2
= cos(x)^2 - ( 1 - cos(x)^2)
= 2cos(x)^2 - 1
= 2(1-sin(x)^2) - 1
= 1 - 2sin(x)^2
Now, let x = y/2. We then have the following equalities
cos(y)
= 2cos(y/2)^2 - 1
= 1 - 2sin(y/2)
rearranging each of these gives a half angle formula.
cos(y/2) = ( [ 1 + cos(y) ] / 2 ) ^1/2
sin(y/2) = ( [ 1 - cos(y) ] / 2 ) ^1/2
These can be verified using the obvious tests ( y=0, y=π, 1=cos(y/2)^2 + sin(y/2)^2). We didn't use sin double angle formula, so that's another fair test
2cos(y/2)sin(y/2)
= 2 ( [ 1 + cos(y) ] / 2 ) ^1/2 ( [ 1 - cos(y) ] / 2 ) ^1/2
= 2 ( [ 1 + cos(y) ] [ 1 - cos(y) ] / [2*2] ) ^1/2
= ( 1 - cos(y)^2 ) ^1/2
= ( sin(y)^2 ) ^ 1/2
= sin(y)
Keep in mind that you have to exhibit some care over the sign of the result; the angle 2π = the angle 0, but 2π/2 != 0/2.
August 2, 2005 1:07 AM
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