1,已知sinA+sinA^2=1那么cosA^2+ cosA^4+ cosA^6=?2,(1+tan1) (1+tan2) (1+tan3) (1+tan4)……(1+tan45)=?3,设6sin3B-(cos2A)^2=6,求A,B是多少?4,已知 0
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![1,已知sinA+sinA^2=1那么cosA^2+ cosA^4+ cosA^6=?2,(1+tan1) (1+tan2) (1+tan3) (1+tan4)……(1+tan45)=?3,设6sin3B-(cos2A)^2=6,求A,B是多少?4,已知 0](/uploads/image/z/5457826-10-6.jpg?t=1%2C%E5%B7%B2%E7%9F%A5sinA%2BsinA%5E2%3D1%E9%82%A3%E4%B9%88cosA%5E2%2B+cosA%5E4%2B+cosA%5E6%3D%3F2%2C%281%2Btan1%29+%281%2Btan2%29+%281%2Btan3%29+%281%2Btan4%29%E2%80%A6%E2%80%A6%281%2Btan45%29%3D%3F3%2C%E8%AE%BE6sin3B-%28cos2A%29%5E2%3D6%2C%E6%B1%82A%2CB%E6%98%AF%E5%A4%9A%E5%B0%91%3F4%2C%E5%B7%B2%E7%9F%A5+0)
1,已知sinA+sinA^2=1那么cosA^2+ cosA^4+ cosA^6=?2,(1+tan1) (1+tan2) (1+tan3) (1+tan4)……(1+tan45)=?3,设6sin3B-(cos2A)^2=6,求A,B是多少?4,已知 0
1,已知sinA+sinA^2=1那么cosA^2+ cosA^4+ cosA^6=?
2,(1+tan1) (1+tan2) (1+tan3) (1+tan4)……(1+tan45)=?
3,设6sin3B-(cos2A)^2=6,求A,B是多少?
4,已知 0
1,已知sinA+sinA^2=1那么cosA^2+ cosA^4+ cosA^6=?2,(1+tan1) (1+tan2) (1+tan3) (1+tan4)……(1+tan45)=?3,设6sin3B-(cos2A)^2=6,求A,B是多少?4,已知 0
1、sinA+sinA^2=1,
=》sinA=(-1+√5)/2,(sinA=(-1-√5)/2舍去)
sinA=1-sin²A=cos²A
cosA^2+ cosA^4+ cosA^6=sinA+sin²A+sin³A
=sinA+sinA(sinA+sin²A)=2sinA=-1+√5
2、1=tan(1°+44°)=(tan1°+tan44°)/(1-tan1°tan44°)
=>tan1°+tan44°+tan1°tan44°+1=2
=>(1+tan1°)(1+tan44°)=2
同理(1+tan2°)(1+tan43°)=2
同理(1+tan22°)(1+tan23°)=2
则(1+tan1)(1+tan2)(1+tan3)(1+tan4)…(1+tan45)=2^23
3、6sin³B-(cos²A)²=6
6sin³B=6+(cos²A)²≥6
则sinB=1,cosA=0
B=2kπ+π/2,A=kπ+π/2(k为任意整数)
4、已知 0lg(1-sin²A)=m-n
=>lgcos²A=m-n
因0
(1)解由已知得sinA=(√5-1)/2,sinA^2=(3-√5)/2
cosA^2=(√5-1)/2
从而代入要求式化简得原式=3-√5.
(2)先证当A+B=45时(1+tanA)(1+tanB)=2
则易证(1+tan1) (1+tan2) (1+tan3) (1+tan4)……(1+tan45)=2^(23)
(4)m-n=lg(1+sinA)(1-sinA)=2lgcosA
lgcosA=(m-n)/2