sina+cosa=1/5、sin²a+cos²a=1,=>sin²a+(1/5-sin²a)=1,
问题描述:
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?
答
(a-b)²=a²-2ab+b²,则:
[sina-(1/5)]²=(sina)²-2×(1/5)×(sina)+(1/5)²=sin²a-(2/5)sina+(1/25)
则:
sin²a-(2/5)sina+(1/25)+sin²a=1
2sin²a-(2/5)sina-(24/25)=0
…………