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Mathematics 14 Online
OpenStudy (anonymous):

Prove that 2sin2Θ−3cos2θ−3sinθ+3=sinθ(4cosθ+6sinθ−3)

OpenStudy (anonymous):

use sin(2x)=2sinxcosx and cos(2x)=1-2sin^2(x) then it'll solve itself easily

OpenStudy (anonymous):

Okay thanks for the help

OpenStudy (anonymous):

yw

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