若sin(a-π/6)=1/3,则cos(2a+π/3)的值为( )
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若sin(a-π/6)=1/3,则cos(2a+π/3)的值为( )
A.√2/2 B.1 C.√2 D.√3
A.√2/2 B.1 C.√2 D.√3
![若sin(a-π/6)=1/3,则cos(2a+π/3)的值为( )](/uploads/image/z/20204071-7-1.jpg?t=%E8%8B%A5sin%28a-%CF%80%2F6%29%3D1%2F3%2C%E5%88%99cos%282a%2B%CF%80%2F3%29%E7%9A%84%E5%80%BC%E4%B8%BA%28+%29)
cos(2a+π/3)=cos[(a-π/6)+(a+π/2)]
cos(a-π/6)=开根号1-sin(a-π/6)*sin(a-π/6)
sin(a-π/6)展开求解
cos(a-π/6)=开根号1-sin(a-π/6)*sin(a-π/6)
sin(a-π/6)展开求解
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