[b]uHalf angle formulae[/u]/b

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 bnot/b provided in the formula book. You don't have to memorise them, as long as you can derive them.

[b]uPower reduction formulae[/u]/b

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 bnot/b 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})

[b]uHalf angle formulae[/u]/b

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

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

[b]uPower reduction formulae[/u]/b

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

(\cos^2{A} ≡ \dfrac{1 + \cos{2A}}{2})