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Trigonometric Identities and Equations, Page 20

Derivation of the Half-Angle Identities

Finally, let’s see how to derive the Half-Angle Identities. You will need the Double-Angle Identities for this process.

To find cosα/2 = ± square root [(1 + cosα)/2]:

cos2θ = 2 (cosθ)^2 - 1

Solve for costheta: 

2 (cosθ)^2 = cos2θ + 1; (cosθ)^2 = (cos2θ + 1)/2; cosθ = ± square root [(cos2θ + 1)/2]

Replace 2θ = α, so therefore:θ = α/2.

cosα/2 = ± square root [(cosα + 1)/2]

To find sinα/2 = ± square root [(1 - cosα)/2]:

cos2θ = 1 - (sinθ )^2

Solve for sintheta: 

-2 (sinθ )^2 = cos2θ - 1; 2 (sinθ )^2 = 1 - cos2θ; (sinθ )^2 = (1 - cos2θ)/2; sinθ = ± square root [(1 - cos2θ)/2]

Again, replace 2θ = α; so therefore: θ = α/2.

sinα/2 = ± square root [(1 - cosα)/2]

To find tanα/2 = ± square root [(1 - cosα)/(1 + cosα)], cosα does not equal -1:

tanα/2 = sin(α/2) / cos(α/2); tanα/2 = (± square root [(1 - cosα)/2])/((± square root [(1 + cosα)/2]) = ± square root [(1 - cosα)/2 * 2/(1 + cosα)] = ± square root [(1 - cosα)/(1 + cosα)]