Một số công thức biến đổi lượng giác

\(\begin{array}{l}\sin (a + b) = \sin a.\cos b + \sin b.\cos a\\\sin (a - b) = \sin a.\cos b - \sin b.\cos a\\\cos (a + b) = \cos a.\cos b - \sin a.\sin b\\\cos (a - b) = \cos a.\cos b + \sin a.\sin b\\\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}\\\tan (a - b) = \dfrac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}\end{array}\)

$\sin 2\alpha  = 2\sin \alpha .\cos \alpha $

\(\cos 2\alpha \,\, = \,\,{\cos ^2}\alpha  - {\sin ^2}\alpha \,\, \)

\(= \,\,2{\cos ^2}\alpha  - 1\,\, \) \(= \,\,1 - 2{\sin ^2}\alpha \)

\(\tan 2\alpha \,\, = \,\,\dfrac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }}\)

\(\begin{array}{l}\sin 3\alpha  = 3\sin \alpha  - 4{\sin ^3}\alpha \\\cos 3\alpha  = 4{\cos ^3}\alpha  - 3\cos \alpha \\\tan 3\alpha  = \dfrac{{3\tan \alpha  - {{\tan }^3}\alpha }}{{1 - 3{{\tan }^2}\alpha }}\end{array}\)

\(\begin{array}{c}{\sin ^2}\alpha \,\, = \,\,\dfrac{{1 - \cos 2\alpha }}{2}\\{\cos ^2}\alpha \, = \,\,\dfrac{{1 + \cos 2\alpha }}{2}\\{\tan ^2}\alpha \, = \,\,\dfrac{{1 - \cos 2\alpha }}{{1 + \cos 2\alpha }}\end{array}\)

\(\begin{array}{l}{\cos ^3}\alpha  = \dfrac{{3\cos \alpha  + \cos 3\alpha }}{4}\\{\sin ^3}\alpha  = \dfrac{{3\sin \alpha  - \sin 3\alpha }}{4}\end{array}\)

$\cos a\cos b $ $= \dfrac{1}{2}\left[ {\cos (a + b) + \cos (a - b)} \right]$

$\sin a\sin b $ $=  - \dfrac{1}{2}\left[ {\cos (a + b) - \cos (a - b)} \right]$

$\sin a\cos b $ $= \dfrac{1}{2}\left[ {\sin (a + b) + \sin (a - b)} \right]$

\(\begin{array}{l}\cos a + \cos b = 2\cos \dfrac{{a + b}}{2}.\cos \dfrac{{a - b}}{2}\\\cos a - \cos b =  - 2\sin \dfrac{{a + b}}{2}.\sin \dfrac{{a - b}}{2}\\\sin a + \sin b = 2\sin \dfrac{{a + b}}{2}.\cos \dfrac{{a - b}}{2}\\\sin a - \sin b = 2\cos \dfrac{{a + b}}{2}.\sin \dfrac{{a - b}}{2}\\\tan a + \tan b = \dfrac{{\sin (a + b)}}{{\cos a.\cos b}}\\\tan a - \tan b = \dfrac{{\sin (a - b)}}{{\cos a.\cos b}}\\\cot a + \cot b = \dfrac{{\sin (a + b)}}{{\sin a.\sin b}}\\\cot a - \cot b = \dfrac{{\sin (b - a)}}{{\sin a.\sin b}}\end{array}\)