\dfrac{x_1+x_2+\dots+x_n}{n} \geq \sqrt[n]{x_1\cdotp x_2 \dots x_n }

and that equality holds if and only if x_1 = x_2 = \dots = x_n

S_n=\dfrac{a_1(1-::r::^n)}{1-::r::}

\sin (\alpha +\beta ) = \sin \alpha\cos \beta +\cos \alpha \sin \beta

\cos(\alpha -\beta ) = \cos \alpha\cos \beta +\sin \alpha \sin \beta

\cos 2\alpha =2 \cos^2 \alpha - 1

\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha

\sin 2\alpha = 2 \sin \alpha\cos \alpha

\tan(\alpha -\beta) = \dfrac{\tan \alpha - \tan \beta }{1+\tan \alpha\tan \beta}

\tan(\alpha+\beta) = \dfrac{\tan \alpha + \tan \beta }{1-\tan \alpha\tan \beta}

\sin \alpha - \sin \beta = 2\cos\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2}

\sin \alpha + \sin \beta = 2\sin\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

\cos \alpha - \cos \beta = -2\sin\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2}

\cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

\cos \alpha\sin \beta = \dfrac{1}{2} [\sin(\alpha+\beta ) - \sin(\alpha - \beta) ]

\sin \alpha\cos \beta = \dfrac{1}{2} [\sin(\alpha+\beta ) +\sin(\alpha - \beta) ]