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时间:2020-03-26
《两种方法证明三角恒等式.pdf》由会员上传分享,免费在线阅读,更多相关内容在应用文档-天天文库。
1、1-2sin2xcos2x1-tan(720°+2x )求证22 =cos2x-sin2x1+tan(360°+2x )分析本题采用“右边Þ左边”和“左边、右边Þ中间”两种方法证明.本题中的角是“2x”,同角三角函数的基本关系式中,内涵在“同角”二字上sin2x1-1-tan2xcos2xcos2x-sin2 x 证法一:右边= ==1+tan2xsin2xcos2x+sin2x 1+cos2x2(cos2x-sin2x ) =(cos2x+sin2x)(cos2x-sin2x )221-2sin2x
2、cos2 x cos2x+sin2x-2cos2xsin2x==22cos22x-sin2 2x cos2x-sin2x=左边.等式成立.证法二:221-2sin2xcos2 x cos2x+sin2x-2cos2xsin2x左边==22 22cos2x-sin2x cos2x-sin2x2 (cos2x-sin2x ) cos2x-sin2 x ==(cos2x+sin2x)(cos2x-sin2x )cos2x+sin2x sin2x1-cos2x1-tan2x===右sin2x1+tan2x1+
3、cos2x∴左边=右边.等式成立.sin2x1-1-tan2xcos2xcos2x-sin2x证法三:右边= ==1+tan2xsin2xcos2x+sin2x1+cos2x221-2sin2xcos2 x cos2x+sin2x-2cos2xsin2x左边==22 22cos2x-sin2x cos2x-sin2x2(cos2x-sin2x ) cos2x-sin2x==(cos2x+sin2x)(cos2x-sin2x )cos2x+sin2x∴左边=右边.等式成立.
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