化简:sin²αsin²β+cos²αcos²β-½cos2αcos2β

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 02:50:17
化简:sin²αsin²β+cos²αcos²β-½cos2αcos2β

化简:sin²αsin²β+cos²αcos²β-½cos2αcos2β
化简:sin²αsin²β+cos²αcos²β-½cos2αcos2β

化简:sin²αsin²β+cos²αcos²β-½cos2αcos2β
sin²αsin²β+cos²αcos²β-(1/2)cos2αcos2β
=sin²αsin²β+cos²αcos²β-(1/2)(1-2sin²α)(1-2sin²β)
=sin²αsin²β+cos²αcos²β-(1/2)(1-2sin²α-2sin²β+4sin²αsin²β)
=sin²αsin²β+cos²αcos²β-2sin²αsin²β+sin²α+sin²β- 1/2
=cos²αcos²β-sin²αsin²β+sin²α+sin²β-1/2
=cos²αcos²β+sin²α(1-sin²β)+sin²β-1/2
=cos²αcos²β+sin²αcos²β+sin²β- 1/2
=(cos²α+sin²α)cos²β+sin²β-1/2
=cos²β+sin²β- 1/2
=1/2

是这个吗
sin2α+sin2β-sin2αsin2β+cos2αcos2β
=sin2α(1-sin2β)+sin2β+cos2αcos2β
=sin2α•cos2β+sin2β+cos2αcos2β
=cos2β(sin2α+cos2α)+sin2β
=cos2β+sin2β=1
故答案为:1