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已知sin(π/3+A)=3/5 则Cos(5π/6+A) 值为?

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已知sin(π/6+ a)=2/3 则sin(7π/6+ a)=sin【π+(π/6+ a)】=-sin(π/6+ a)=-2/3

阿尔法减贝塔=三分之丌

6π÷2π=3,是2π的倍数, 所以cosα=cos(6π+α)=-√[1-sin(6π+α)]²= -4/5 tanα=tan(6π+α)=sin(6π+α) /cos(6π+α) =(3/5 )/(-4/5)=-3/4

由 cos(α+π/6)=3/5,得sin(α+π/6)=4/5 cos(2a-π/6) =cos(2a+π/3-π/2) =-cos【π/2-(2a+π/3)】 =-sin(2a+π/3) =-sin2(a+π/6) =-2sin(a+π/6)cos(a+π/6) =-2x4/5x3/5 =-24/25 谢谢,请采纳

是不是题错了?是不是cos(π/6-a)=根号3 除以2??? cos(5/6π+a)=-cos(π-5/6π-a)=-cos(π/6-a)=-(根号3)除以2 sin(a+π/6)=sin(a-π/6+π/3)=sin(a-π/6)cosπ/3 + cos(a-π/6)sinπ/3 =1/2 乘以1/2+ 【(根号3) 除以2】²=1 或者=-1/2 ...

解:∵cos(a+π/6)-sina=5√3 ==>sin(π/2-a-π/6)-sina=5√3 (应用诱导公式) ==>sin(π/3-a)-sina=5√3 ==>2cos(π/6)sin(π/6-a)=5√3 (应用和差化积公式) ==>√3sin(π/6-a)=5√3 ==>sin(π/6-a)=5 ∴sin(a+5π/6)=sin(π-(π/6-a)) =sin(π/6-a) (应用诱导公式) ...

sin(a-5π/3)=sin(a-5π/3+2π)=sin(a+π/3)=-1/2

题干不详

cosα=3/5,0

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