If we rewrite α as α/2 cos(2·α/2)=1-2·sin^{2}(α/2)-1 As: cos(2·α/2)=cos(α), We have: cos(α)=1−2·sin^{2}(α/2) [1.2] By rearranging : 2·sin^{2}(α/2)=1-cos(α)

So: sin^{2}(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·cos^{2}α=cos2α+1

Rewriting α as α/2: 2·cos^{2}(α)/2=cosα+1 Rearranging: 2cos^{2}(α/2)=cosα+1 cos^{2}(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-cos^{2}α)) As 1-cos^{2}α=sin^{2}α, we have: = √((1-cosα)^{2}/(sin^{2}α)) Taking the square root: = ±((1-cosα)/(sinα)) [1.8] or [1.9] ■

Dividing by cos^{2}(α/2)·sec^{2}(α/2) (which is equal to 1): sin(α)=2·tan(α/2)/sec^{2}(α/2) =2·tan(α/2)/1+tan^{2}(α/2) Because sec^{2}(α/2)=1+tan^{2}(α/2)

Divide by cos^{2}(α/2)·sec^{2}(α/2) (which is equal to 1): cos(α)=[sec^{2}(α/2)−2·tan^{2}(α/2)]/sec^{2}(α/2)

We note sec^{2}(α/2)=1+tan^{2}(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+t^{2}), base, b=(1-t^{2}), and perpendicular, p=2t, so h^{2}=b^{2}+p^{2},

and substituting the half tangent, t, values, we get: (1+t^{2})^{2}=(1-t^{2})^{2}+(2t)^{2} ■

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: