Question

化简：sin²αsin²β+cos²αcos²β-½cos2αcos2β

Answer 1

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

Answer 2

是这个吗
解：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