数学(二)必背公式[基础知识] 因式分解公式:a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})(n为正偶数时(1)a^{n}-b^{n}=(a+b)(a^{n-1}-a^{n-2}b+\cdots+ab^{n-2}-b^{n-1})(n为正奇数时)a^{n}+b^{n}=(a+b)(a^{n-1}-a^{n-2}b+\cdots -ab^{n-2}+b^{n-1})二项式定理:[(a+b)^{n}= \sum _{k=0}^{n}C_{n}^{k}a^{k}b^{n-k} 不等式：(1)a,b位实数,则\circled {4}2|ab| \le a^{2}+b^{2};\circled {2}|a \pm b| \le |a|+|b|;\circled {3}|a|-|b| \le |a-b|.(2)a_{1},a_{2}, \cdots ,a_{n}>0, 则 \frac {a_{1}+a_{2}+\cdots+a_{n}}{n} \ge \sqrt [n]{a_{1}a_{2} \cdots a_{n}} 取整函数:\mid x-1<[x] \le x 三角函数和差化积;积化和差(7):\sin \alpha+\sin \beta =2(\sin \frac { \alpha+\beta }{2})(\cos \frac { \alpha - \beta }{2})\sin \alpha \cos \beta = \frac {1}{2}(\sin \frac { \alpha+\beta }{2}+\cos \frac { \alpha - \beta }{2})\sin \alpha - \sin \beta =2(\cos \frac { \alpha+\beta }{2})(\sin \frac { \alpha - \beta }{2})\cos \alpha \cos \beta = \frac {1}{2}(\cos \frac { \alpha+\beta }{2}+\cos \frac { \alpha - \beta }{2})\cos \alpha+\cos \beta =2(\cos \frac { \alpha+\beta }{2})(\cos \frac { \alpha - \beta }{2})\sin \alpha \sin \beta =- \frac {1}{2}(\cos \frac { \alpha+\beta }{2}