1.cos(π/6 - a)=根号3 /3,求cos(5π/6+a)+sin^2(a-π/6)的值2.已知sinA,sinB是方程8x^2-6kx+2k+1=0的两个根,且A、B角的终边互相垂直,求实数k的值.
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![1.cos(π/6 - a)=根号3 /3,求cos(5π/6+a)+sin^2(a-π/6)的值2.已知sinA,sinB是方程8x^2-6kx+2k+1=0的两个根,且A、B角的终边互相垂直,求实数k的值.](/uploads/image/z/13902107-59-7.jpg?t=1.cos%28%CF%80%2F6+-+a%29%3D%E6%A0%B9%E5%8F%B73+%2F3%2C%E6%B1%82cos%285%CF%80%2F6%2Ba%29%2Bsin%5E2%28a-%CF%80%2F6%29%E7%9A%84%E5%80%BC2.%E5%B7%B2%E7%9F%A5sinA%2CsinB%E6%98%AF%E6%96%B9%E7%A8%8B8x%5E2-6kx%2B2k%2B1%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%2C%E4%B8%94A%E3%80%81B%E8%A7%92%E7%9A%84%E7%BB%88%E8%BE%B9%E4%BA%92%E7%9B%B8%E5%9E%82%E7%9B%B4%2C%E6%B1%82%E5%AE%9E%E6%95%B0k%E7%9A%84%E5%80%BC.)
1.cos(π/6 - a)=根号3 /3,求cos(5π/6+a)+sin^2(a-π/6)的值2.已知sinA,sinB是方程8x^2-6kx+2k+1=0的两个根,且A、B角的终边互相垂直,求实数k的值.
1.cos(π/6 - a)=根号3 /3,求cos(5π/6+a)+sin^2(a-π/6)的值
2.已知sinA,sinB是方程8x^2-6kx+2k+1=0的两个根,且A、B角的终边互相垂直,求实数k的值.
1.cos(π/6 - a)=根号3 /3,求cos(5π/6+a)+sin^2(a-π/6)的值2.已知sinA,sinB是方程8x^2-6kx+2k+1=0的两个根,且A、B角的终边互相垂直,求实数k的值.
.cos(π/6 - a)=根号3 /3,求cos(5π/6+a)+sin^2(a-π/6)的值
cos(5∏/6+a)=cos[∏-(∏/6-a)]=-cos(∏/6-a)=-根号3/3
由公式 ( sina)^2+(cosa)^2=1
sin^2(π/6-a)=1-cos^2(π/6-a)=1-(根号3/3)^2=1-1/3=2/3
故,原式=-根号3/3+2/3
2.已知sinA,sinB是方程8x^2-6kx+2k+1=0的两个根,且A、B角的终边互相垂直,求实数k的值.
设A=α,则B=90+α,sinA=sinα,sinB=sin(90+α)=cosα
由韦达定理可得:
sinα+cosα=3k/4
sinα*cosα=(2k+1)/8
又因为;(sinα)^2+(cosα)^2=1
则:(sinα+cosα)^2-2sinα*cosα=1
代入得:(3k/4)^2-2[(2k+1)/8]=1
解得:k=2或k=-10/9
代入判别式得:k=-10/9