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sina+cosa=1/5、sin²a+cos²a=1,=>sin²a+(1/5-sin&
题目内容:
sina+cosa=1/5、sin²a+cos²a=1,=>sin²a+(1/5-sin²a)=1,
=>sin²a+(1/5-sin²a)=1
sin²a-(2/5)sina+(1/25)+sin²a=1
2sin²a-(2/5)sina-(24/25)=0
25sin²a-5sina-12=0
(5sina+3)(5sina-4)=0
sina=-3/5【舍去】,sina=4/5
问:2sin²a-(2/5)sina-(24/25)=0这步中的2/5是怎么得的?
sina+cosa=1/5、sin²a+cos²a=1,=>sin²a+(1/5-sin²a)=1,
=>sin²a+(1/5-sin²a)=1
sin²a-(2/5)sina+(1/25)+sin²a=1
2sin²a-(2/5)sina-(24/25)=0
25sin²a-5sina-12=0
(5sina+3)(5sina-4)=0
sina=-3/5【舍去】,sina=4/5
问:2sin²a-(2/5)sina-(24/25)=0这步中的2/5是怎么得的?
=>sin²a+(1/5-sin²a)=1
sin²a-(2/5)sina+(1/25)+sin²a=1
2sin²a-(2/5)sina-(24/25)=0
25sin²a-5sina-12=0
(5sina+3)(5sina-4)=0
sina=-3/5【舍去】,sina=4/5
问:2sin²a-(2/5)sina-(24/25)=0这步中的2/5是怎么得的?
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