已知a属于(0,π),cos(π/3+a)=1/3,求sin(π/3-2a),
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已知a属于(0,π),cos(π/3+a)=1/3,求sin(π/3-2a),
![已知a属于(0,π),cos(π/3+a)=1/3,求sin(π/3-2a),](/uploads/image/z/16405970-50-0.jpg?t=%E5%B7%B2%E7%9F%A5a%E5%B1%9E%E4%BA%8E%EF%BC%880%2C%CF%80%EF%BC%89%2Ccos%28%CF%80%2F3%2Ba%29%3D1%2F3%2C%E6%B1%82sin%EF%BC%88%CF%80%2F3-2a%29%2C)
cos(π/3+a)=1/3
→cos^2 (π/3+a)=1/9
→2cos^2 (π/3+a)-1=-7/9;
即cos(2π/3+2a)=-7/9;
则cos(-2π/3-2a)=-7/9;
则cos[(-2π/3-2a)+π]=+7/9;
即cos(π/3-2a)=7/9;
a∈(0,π),则
π/3+a∈(π/3,4π/3);
而cos(π/3+a)=1/3>0,
则说明π/3<π/3+a<π/2.
→0<a<π/6;
→π/3-2a∈(0,π/3);
则sin(π/3-2a)>0;
sin(π/3-2a)=√[1-cos^2 (π/3+a)]
=(2√2)/3
→cos^2 (π/3+a)=1/9
→2cos^2 (π/3+a)-1=-7/9;
即cos(2π/3+2a)=-7/9;
则cos(-2π/3-2a)=-7/9;
则cos[(-2π/3-2a)+π]=+7/9;
即cos(π/3-2a)=7/9;
a∈(0,π),则
π/3+a∈(π/3,4π/3);
而cos(π/3+a)=1/3>0,
则说明π/3<π/3+a<π/2.
→0<a<π/6;
→π/3-2a∈(0,π/3);
则sin(π/3-2a)>0;
sin(π/3-2a)=√[1-cos^2 (π/3+a)]
=(2√2)/3
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