Question

化简sin²αsin²β +cos²αcos²β-1/2cos2αcosβ

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Answer 1

Sin²αSin²β+Cos²αCos²β-1/2Cos2αCos2β
=Sin²αSin²β+Cos²αCos²β-1/2(Cos²α-Sin²α)(Cos²β-Sin²β)
=Sin²αSin²β+Cos²αCos²β-1/2(Cos²αCos²β-Cos²αSin²β-Sin²αCos²β+Sin²αSin²β)
=1/2(Cos²αCos²β+Cos²αSin²β+Sin²αCos²β+Sin²αSin²β)
=1/2(Cos²α+Sin²α)(Cos²β+Sin²β)
=1/2