若sin(π/6-x)=1/3,则cos(2π/3+2x)=___________
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若sin(π/6-x)=1/3,则cos(2π/3+2x)=___________
![若sin(π/6-x)=1/3,则cos(2π/3+2x)=___________](/uploads/image/z/7791388-52-8.jpg?t=%E8%8B%A5sin%28%CF%80%2F6-x%29%3D1%2F3%2C%E5%88%99cos%282%CF%80%2F3%2B2x%29%3D___________)
cos(2π/3+2x)
=cos[π-(π/3-2x)]
=-cos(π/3-2x)
=-cos2(π/6-x)
=-[1-2sin²(π/6-x)]
=-[1-2*(1/3)²]
=-(1-2/9)
=-7/9
=cos[π-(π/3-2x)]
=-cos(π/3-2x)
=-cos2(π/6-x)
=-[1-2sin²(π/6-x)]
=-[1-2*(1/3)²]
=-(1-2/9)
=-7/9
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