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sin a sin b +cos a cos b =cos(a-b)=0
所以a-b=π/2+kπ 所以a=π/2+kπ +b 所以2a=π+2kπ +2b
sin a cos a+sin b cos b=1/2sin(2a)+1/2sin(2b)
=1/2sin(π+2kπ +2b )+1/2sin(2b)
=1/2sin(π +2b )+1/2sin(2b)
=-1/2sin(2b )+1/2sin(2b)
=0
所以a-b=π/2+kπ 所以a=π/2+kπ +b 所以2a=π+2kπ +2b
sin a cos a+sin b cos b=1/2sin(2a)+1/2sin(2b)
=1/2sin(π+2kπ +2b )+1/2sin(2b)
=1/2sin(π +2b )+1/2sin(2b)
=-1/2sin(2b )+1/2sin(2b)
=0
1年前
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