如果a∈(π⼀2,π)且sina=4⼀5,那么sin(a+π⼀4)+cos(a=π⼀4)=

2025-12-17 23:32:04
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回答1:

a∈(π/2,π)且sina=4/5,所以cosa=-3/5
sin(a+π/4)+cos(a+π/4)
=sinacosπ/4+cosasinπ/4+cosacosπ/4-sinasinπ/4
=[(根号2)/2](sina+cosa+cosa-sina)
=(根号2)cosa
=-3(根号2)/5