若cos(a-π/3)=3/5 则sin(2a-π/6)=

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若cos(a-π/3)=3/5 则sin(2a-π/6)=
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若cos(a-π/3)=3/5 则sin(2a-π/6)=
若cos(a-π/3)=3/5 则sin(2a-π/6)=

若cos(a-π/3)=3/5 则sin(2a-π/6)=
sin(a-π/3)=√[1-cos^2(a-π/3)]=4/5
sin(2a-π/6)=sin2(a-π/3)=2sin(a-π/3)cos(a-π/3)=2x3/5x4/5=24/25

令t=a-π/3,则2a-π/6=2t
所以cost=3/5,sint=√(1-(cost)^2)=4/5
sin(2a-π/6)=sin2t=2sint*cost=2*3/5*4/5=24/25大哥 谢谢你回答 但是你看清题啊 是2a-π/6与a-π/3 不是2倍的关系啊 2t=2a-2π/3奥,不好意思啊!
sin(2a...

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令t=a-π/3,则2a-π/6=2t
所以cost=3/5,sint=√(1-(cost)^2)=4/5
sin(2a-π/6)=sin2t=2sint*cost=2*3/5*4/5=24/25

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