Trigonometric addition formula

Formulas

\(\begin{align} \cos(a-b) = \cos{a}\cos{b} + \sin{a}\sin{b} \\ \cos(a+b) = \cos{a}\cos{b} - \sin{a}\sin{b} \\ \sin(a+b) = \sin{a}\cos{b} + \cos{a}\sin{b} \\ \cos(a-b) = \sin{a}\cos{b} - \cos{a}\sin{b} \end{align}\)

Derivation

\[\begin{flalign*} \\ \overline{AB}^2 &= (\cos{a}-\cos{b})^2 + (\sin{a}-\sin{b})^2 \\ &= (\cos^2a + \sin^2a) + (\cos^2b + \sin^2b) -2\cos{a}\cos{b}-2\sin{a}\sin{b} \\ &= 2 - 2\cos{a}\cos{b}-2\sin{a}\sin{b} \\ && \end{flalign*}\]

and

\[\begin{flalign*} \overline{AB}^2 &= \overline{OA}^2 + \overline{OB}^2 -2\overline{OA}\ \overline{OB}\cos(a-b) \\ &= 2 - \cos(a-b) \\ && \end{flalign*}\]

By solving the system of equations, you can derive the formula as follows.

\[\begin{flalign*} \cos(a-b) = \cos{a}\cos{b} + \sin{a}\sin{b} \\ && \end{flalign*}\]

By applying this formula, you can derive other equations.

\[\begin{flalign*} \cos(a+b) &= \cos{a}\cos(-b)+\sin{a}\sin(-b) \\ &= \cos{a}\cos{b}-\sin{a}\sin{b} \\ && \end{flalign*}\]

Also, from the relations between sine and cosine,

\(\begin{flalign*} \sin{x} = \cos(x-\pi/2) && \end{flalign*}\) and \(\begin{flalign*} \cos{x} = \sin(x-\pi/2) \\ && \end{flalign*}\)

you can derive following formula.

\[\begin{flalign*} \sin(a+b) &= \cos(a+b-\pi/2) \\ &= \cos a\cos(b-\pi/2)-\sin a\sin(b-\pi/2) \\ &= \sin{a}\cos{b}+\cos{a}\sin{b} \\ && \end{flalign*}\]

Final formula can be derived as follows.

\[\begin{flalign*} \sin(a-b) &= \sin a\cos(-b)+\cos a\sin(-b) \\ &= \sin{a}\cos{b}-\cos{a}\sin{b} \\ && \end{flalign*}\]