tanalpha -tanbeta =tan(alpha -beta )(1+tanalpha tanbeta ),综上所述,答案是:tan(alpha pm beta )(1stackrel - +tanalpha tanbeta );(2)由cos2alpha =2(cos)^2alpha -1=1-2(sin)^2alpha ,得:(cos)^2alpha =dfrac (1+cos2alpha ) 2,(sin)^2alpha =dfrac (1-cos2alpha ) 2,综上...
百度试题 结果1 题目【题目】【题目】公式 _ 以变形为 _ \$\tan \beta = \tan ( \alpha + \beta ) ( 1 - \tan \alpha \tan \beta )\$ ,且对任意角 _ 都成立。 相关知识点: 试题来源: 解析 【解析】× 反馈 收藏
(\$ 3.公式的逆用、变形等 \$\tan \alpha \pm \tan \beta = \tan ( \alpha \pm \beta ) (\$ 3.公式的逆用、变形等 \$\tan \alpha \pm \tan \beta = \tan ( \alpha \pm \beta ) (\$ 3.公式的逆用、变形等 \$\tan \alpha \pm \tan \beta = \tan ( \alpha \pm \beta )...