利用公式求下列三角函数値: (l). cos(-420°);(2). sin(-7/6π);(3). sin(-1300°);(4). cos(-79/6π) 30
化简(1)sin(α+180°)cos(-α)sin(-α-180°)(2)sin^3(-α)cos(2π+α)tan(-α-π)要过程啊!!!!急啊,快啊!!!...
化简(1)sin(α+180°)cos(-α)sin(-α-180°)
(2)sin^3(-α)cos(2π+α)tan(-α-π)要过程啊!!!!急啊,快啊!!! 展开
(2)sin^3(-α)cos(2π+α)tan(-α-π)要过程啊!!!!急啊,快啊!!! 展开
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1,(1) cos(-420°)=cos(-420+360)=cos(-60)=cos60=1/2
(2)sin(-7/6π)=sin(-7π/6+2π)=sin5π/6=sin(π-5π/6)=sinπ/6=1/2
(3)sin(-1300°)=sin(-1300+360*4)=sin140=sin40
(4) cos(-79π/6)=cos(-79π/6+2π*7)=cos(5π/6)=-cosπ/6=-根号3/2
2.(1)sin(α+180°)cos(-α)sin(-α-180°)=-sinα*cosα*sinα=-cosα*(sinα)^2
(2)sin^3(-α)cos(2π+α)tan(-α-π)=-(sinα)^3*cosα*(-tanα)=(sinα)^4
(2)sin(-7/6π)=sin(-7π/6+2π)=sin5π/6=sin(π-5π/6)=sinπ/6=1/2
(3)sin(-1300°)=sin(-1300+360*4)=sin140=sin40
(4) cos(-79π/6)=cos(-79π/6+2π*7)=cos(5π/6)=-cosπ/6=-根号3/2
2.(1)sin(α+180°)cos(-α)sin(-α-180°)=-sinα*cosα*sinα=-cosα*(sinα)^2
(2)sin^3(-α)cos(2π+α)tan(-α-π)=-(sinα)^3*cosα*(-tanα)=(sinα)^4
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