Question

(sinα)^2*(sinβ)^2+(cosα)^2(cosβ)^2-1/2cos2αcos2β

各位亲啊！！这个题是来自哪里的啊？离啊

Answer 1

(sinα)^2*(sinβ)^2+(cosα)^2(cosβ)^2-1/2cos2αcos2β
=sin²αsin²β+cos²αcos²β-(cos2αcos2β)/2
=sin²αsin²β+cos²αcos²β-(1-2sin²α)(1-2sin²β)/2
=sin²αsin²β+cos²αcos²β-(1-2sin²α-2sin²β+4sin²αsin²β)/2
=sin²αsin²β+cos²αcos²β-1/2+sin²α+sin²β-2sin²αsin²β
=cos²αcos²β+sin²α+sin²β+sin²αsin²β-2sin²αsin²β-1/2
=(1-sin²α)(1-sin²β)+sin²α+sin²β-sin²αsin²β-1/2
=1-sin²α-sin²β+sin²αsin²β+sin²α+sin²β-sin²αsin²β-1/2
=1-sin²α-sin²β+sin²α+sin²β+sin²αsin²β-sin²αsin²β-1/2
=1-1/2
=1/2

追问

请问这道题是在哪的。是哪年的高考题吗？还是书上的内容啊？