已知cos(a-π/6)=√3 /3,求sin(a-π/6)-cos(5π/6+a)的值

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已知cos(a-π/6)=√3/3,求sin(a-π/6)-cos(5π/6+a)的值已知cos(a-π/6)=√3/3,求sin(a-π/6)-cos(5π/6+a)的值已知cos(a-π/6)=√

已知cos(a-π/6)=√3 /3,求sin(a-π/6)-cos(5π/6+a)的值
已知cos(a-π/6)=√3 /3,求sin(a-π/6)-cos(5π/6+a)的值

已知cos(a-π/6)=√3 /3,求sin(a-π/6)-cos(5π/6+a)的值
cos(5/6*π+a) =-cos[π-(5/6*π+a)] =-cos[π/6-a] [sin(a-π/6)]^2=-[cos(a-π/6)]^2-1 -cos(π/6-a)=-cos(a-π/6) 所以 -cos(5/6π+a)+[sin(a-π/6)]^2 =√3/3-1/3+1 = (2+√3)/3

sin平方(a-pi/6)=1-cos平方(a-pi/6)=1-1/3=2/3 cos(5pi/6+a)=cos(a-pi/6+pi)=cos(a-pi/6)=三分之根三 所以原式等于三分之二减掉三分之根三咯~~~