Half angle formulae

Substituting \(\dfrac{\theta}{2}\) into \(A\) for \(\cos{2A}\) gives:

\(\cos{\theta} ≡ 1 - 2\sin^2{\dfrac{\theta}{2}}\)

\(\implies \sin^2{\dfrac{\theta}{2}} ≡ \dfrac{1 - \cos{\theta}}{2}\)

\(\implies \boxed{\sin{\dfrac{\theta}{2}} ≡ ±\sqrt{\dfrac{1 - \cos{\theta}}{2}}}\)

Alternatively:

\(\cos{\theta} ≡ 2\cos^2{\dfrac{\theta}{2}} - 1\)

\(\implies \dfrac{\cos{\theta} + 1}{2} ≡ \cos^2{\dfrac{\theta}{2}} \)

\(\implies \boxed{±\sqrt{\dfrac{\cos{\theta} + 1}{2}} ≡ \cos{\dfrac{\theta}{2}}} \)

Applying the identity: \(\tan{\theta} ≡ \dfrac{\sin{\theta}}{\cos{\theta}}\)

\(\tan{\dfrac{\theta}{2}} ≡ \dfrac{±\sqrt{\dfrac{1 - \cos{\theta}}{2}}}{±\sqrt{\dfrac{\cos{\theta} + 1}{2}}} \)

\(\tan{\dfrac{\theta}{2}} ≡ ±\sqrt{\dfrac{\dfrac{1 - \cos{\theta}}{\cancel{2}}}{\dfrac{\cos{\theta} + 1}{\cancel{2}}}} \)

\(\implies \boxed{\tan{\dfrac{\theta}{2}} ≡ ±\sqrt{\dfrac{1 - \cos{\theta}}{\cos{\theta} + 1}}} \)

Knowledge of \(\tan{\dfrac{\theta}{2}}\) is not required, but is provided here for completeness.

These are not provided in the formula book. You don't have to memorise them, as long as you can derive them.

Power reduction formulae

Rearranging the identity for \(\cos{2A}\) gives:

\(\cos{2A} ≡ 1 - 2\sin^2{A}\)

\(\implies 2\sin^2{A} ≡ 1 - \cos{2A}\)

\(\implies \boxed{\sin^2{A} ≡ \dfrac{1 - \cos{2A}}{2}} \)

Alternatively:

\(\cos{2A} ≡ 2\cos^2{A} - 1\)

\(\implies \cos{2A} + 1 ≡ 2\cos^2{A}\)

\(\implies \boxed{\dfrac{\cos{2A} + 1}{2} ≡ \cos^2{A}}\)

These are not provided in the formula book. You don't have to memorise them, as long as you can derive them.

Tip: Power reduction formulae are particularly useful for integrating \(\sin^2{A}\) and \(\cos^2{A}\)