The compound angle formulae can be derived from the diagram below:

mtaimg/images/topics/5/5-56-1.png/mtaimg

\implies \boxed{\sin{(A+B)} ≡ \sin{A}\cos{B} + \cos{A}\sin{B}}

\implies \boxed{\cos{(A+B)} ≡ \cos{A}\cos{B} - \sin{A}\sin{B}}

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

\implies \tan{(A+B)} ≡ \dfrac{\sin{A}\cos{B} + \cos{A}\sin{B}}{\cos{A}\cos{B} - \sin{A}\sin{B}}

Dividing all terms by \cos{A}\cos{B}:

\implies \tan{(A+B)} ≡ \dfrac{\dfrac{\sin{A}\cancel{\cos{B}}}{\cos{A}\cancel{\cos{B}}} + \dfrac{\cancel{\cos{A}}\sin{B}}{\cancel{\cos{A}}\cos{B}}} {\dfrac{\cancel{\cos{A}\cos{B}}}{\cancel{\cos{A}\cos{B}}} - \dfrac{\sin{A}\sin{B}}{\cos{A}\cos{B}}}

\implies \boxed{\tan{(A+B)} ≡ \dfrac{\tan{A} + \tan{B}}{1 - \tan{A}\tan{B}}}

Substituting -B into B gives the negative versions:

\implies \boxed{\sin{(A-B)} ≡ \sin{A}\cos{B} - \cos{A}\sin{B}}

\implies \boxed{\cos{(A-B)} ≡ \cos{A}\cos{B} + \sin{A}\sin{B}}

\implies \boxed{\tan{(A-B)} ≡ \dfrac{\tan{A} - \tan{B}}{1 + \tan{A}\tan{B}}}

These are provided in the formula book.

[b]uDouble angle formulae[/u]/b

Substituting A into B for the addition formulae gives the double angle formulae:

\sin{(A+A)} ≡ \sin{A}\cos{A} + \cos{A}\sin{A}

\implies \boxed{\sin{2A} ≡ 2\sin{A}\cos{A}}

\cos{(A+A)} ≡ \cos{A}\cos{A} - \sin{A}\sin{A}

\implies \boxed{\cos{2A} ≡ \cos^2{A} - \sin^2{A}}

\tan{(A+B)} ≡ \dfrac{\tan{A} + \tan{B}}{1 - \tan{A}\tan{B}}

\implies \boxed{\tan{2A} ≡ \dfrac{2\tan{A}}{1 - \tan^2{A}}}

Applying the identity \sin^{2{\theta}+\cos2{\theta}=1} to \cos{2A} provides two further forms:

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

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

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

\implies \boxed{\cos{2A} ≡ 2\cos^2{A} - 1}

These are provided in the formula book.

[b]uCompound angle formulae[/u]/b

\sin{(A±B)} ≡ \sin{A}\cos{B} ± \cos{A}\sin{B}

\cos{(A±B)} ≡ \cos{A}\cos{B} ∓ \sin{A}\sin{B}

\tan{(A±B)} ≡ \dfrac{\tan{A} ± \tan{B}}{1 ∓ \tan{A}\tan{B}}