If we rewrite α as α/2 cos(2·α/2)=1-2·sin2(α/2)-1 As: cos(2·α/2)=cos(α), We have: cos(α)=1−2·sin2(α/2) [1.2] By rearranging : 2·sin2(α/2)=1-cos(α)
So: sin2(α/2)=(1-cos(α))/2
And, by taking the square root, the sine half alpha is: [1.3] or [1.4] ■
Cosine
We again start with the double angle for cosine, but this time use this version: [1.5]
So, by rearranging, 2·cos2α=cos2α+1
Rewriting α as α/2: 2·cos2(α)/2=cosα+1 Rearranging: 2cos2(α/2)=cosα+1 cos2(α/2)=(cosα+1)/2 And by taking the square root, we have two formulae (one for the + and one for the minus root): [1.6]
and [1.7] ■
Tangent
To get the tangent of half an angle, knowing the angle, we divide the sine by the cosine formulae: tan(α/2)=sin(α/2)/cos(α/2)
=√((1-cosα)/(1+cosα))
Multiply top and bottom by 1-cosα =√((1-cosα)2/(1-cos2α)) As 1-cos2α=sin2α, we have: = √((1-cosα)2/(sin2α)) Taking the square root: = ±((1-cosα)/(sinα)) [1.8] or [1.9] ■
Divide by cos2(α/2)·sec2(α/2) (which is equal to 1): cos(α)=[sec2(α/2)−2·tan2(α/2)]/sec2(α/2)
We note sec2(α/2)=1+tan2(α/2), so [2.2] ■
Tangent
Dividing sinα by cosα (Equation 2.1 by 2.2) gives us: [2.3] ■ Alternatively, we note the tangent double angle formula is:
By setting α as α/2, we immediately get the half angle formula (Equation 2.3) ■
Relationship Between Tangent of Half Angles
The
three values that occur in the half tangent formula are sides of a
right angled triangle, so writing t=tan(α/2), and the hypotenuse, h=(1+t2), base, b=(1-t2), and perpendicular, p=2t, so h2=b2+p2,
and substituting the half tangent, t, values, we get: (1+t2)2=(1-t2)2+(2t)2 ■
[1.1]
[1.3]
[1.4]
[1.5]
[1.6]
[1.7]
[1.8]
[1.9]
[2.1]
[2.2]
[2.3]