设cos460°=t则tan260°=_________(用t表示)设cos460°=t,则tan260°=_________(用t表示)化简:(1)根号下1+2sin(π-2)cos(π-2)(2)α是第二象限角,cosα根号下(1-sinα/1+sinα)+sinα根号下(1-cosα/1+cosα)=(
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![设cos460°=t则tan260°=_________(用t表示)设cos460°=t,则tan260°=_________(用t表示)化简:(1)根号下1+2sin(π-2)cos(π-2)(2)α是第二象限角,cosα根号下(1-sinα/1+sinα)+sinα根号下(1-cosα/1+cosα)=(](/uploads/image/z/12518581-13-1.jpg?t=%E8%AE%BEcos460%C2%B0%3Dt%E5%88%99tan260%C2%B0%3D_________%28%E7%94%A8t%E8%A1%A8%E7%A4%BA%29%E8%AE%BEcos460%C2%B0%3Dt%2C%E5%88%99tan260%C2%B0%3D_________%28%E7%94%A8t%E8%A1%A8%E7%A4%BA%29%E5%8C%96%E7%AE%80%EF%BC%9A%EF%BC%881%EF%BC%89%E6%A0%B9%E5%8F%B7%E4%B8%8B1%2B2sin%EF%BC%88%CF%80-2%EF%BC%89cos%EF%BC%88%CF%80-2%EF%BC%89%EF%BC%882%EF%BC%89%CE%B1%E6%98%AF%E7%AC%AC%E4%BA%8C%E8%B1%A1%E9%99%90%E8%A7%92%2Ccos%CE%B1%E6%A0%B9%E5%8F%B7%E4%B8%8B%EF%BC%881-sin%CE%B1%2F1%2Bsin%CE%B1%EF%BC%89%2Bsin%CE%B1%E6%A0%B9%E5%8F%B7%E4%B8%8B%EF%BC%881-cos%CE%B1%2F1%2Bcos%CE%B1%EF%BC%89%3D%EF%BC%88)
设cos460°=t则tan260°=_________(用t表示)设cos460°=t,则tan260°=_________(用t表示)化简:(1)根号下1+2sin(π-2)cos(π-2)(2)α是第二象限角,cosα根号下(1-sinα/1+sinα)+sinα根号下(1-cosα/1+cosα)=(
设cos460°=t则tan260°=_________(用t表示)
设cos460°=t,则tan260°=_________(用t表示)
化简:(1)根号下1+2sin(π-2)cos(π-2)
(2)α是第二象限角,cosα根号下(1-sinα/1+sinα)+sinα根号下(1-cosα/1+cosα)=
(3)cotα根号下(1-cos²α)
已知tanα=3,求值:(1):(4sinα-2cosα)/5cosα+3sinα.(2)sinαcosα.(3)2sin²α+cosαsinα-3cos²α.(4)(sinα+cosα)²
设cos460°=t则tan260°=_________(用t表示)设cos460°=t,则tan260°=_________(用t表示)化简:(1)根号下1+2sin(π-2)cos(π-2)(2)α是第二象限角,cosα根号下(1-sinα/1+sinα)+sinα根号下(1-cosα/1+cosα)=(
∵cos460=t,即cos(360+100)=t
∴cos100=t
∴sin100=根号下(1-cos100的平方)=根号下(1-t^2)
(注:sin100在第二象限,符号为正)
∴tan100=sin100/cos100=(根号下1-t^2)/t
∴tan260=tan(360-100)=-tan 100=-(根号下1-t^2)/t
根号1+2sin(π-2)cos(π-2)
=根号(sin²(π-2)+cos²(π-2)+2sin(π-2)cos(π-2))
=根号(sin(π-2)+cos(π-2))²
=sin(π-2)+cos(π-2)=sin2-cos2
α第二象限角
sinα>0,cosα<0
cosα * 根号[(1-sinα)/(1+sinα)] + sinα *根号〔(1-cosα)/(1+cosα)〕
=cosα * 根号[(1-sinα)^2/(1-sin^2α)] + sinα *根号〔(1-cosα)^2/(1-cos^2α)〕
=cosα * 根号[(1-sinα)^2/cos^2α] + sinα *根号〔(1-cosα)^2/sin^2α〕
=cosα * (1-sinα)/(-cosα) + sinα *(1-cosα)/sinα
=- (1-sinα) + (1-cosα)
=sinα-cosα
(2)cotα根号下(1-cos²α)=cosa
4.∵tanα=3
∴sinα=3cosα.(1)
==>(3cosα)²+(cosα)²=1
==>cos²α=1/10.(2)
故(1)(4sinα-2cosα)/(5cosα+3sinα)
=(4(3cosα)-2cosα)/(5cosα+3(3cosα)) (由(1)得)
=(10cosα)/(14cosα)
=5/7;
(2)sinαcosα
=(3cosα)cosα (由(1)得)
=3cos²α
=3*(1/10) (由(2)得)
=3/10;
(3)2sin²α+cosαsinα-3cos²α
=2(sin²α+cos²α) + cosαsinα-5cos²α
=2+3/10-1/2
=9/5
(4)(sinα+cosα)²
=sin²α+2sinαcosα+cos²α
=1+2*(3/10) (由(2)题得)
=8/5.