Ai giup minh voi nao`

N

nhock.l0ve_95

[TEX]\sqrt{2}cos3x=sinx+cosx [/TEX]
\Leftrightarrow [TEX]\sqrt{2}[/TEX]cos3x.1/[TEX]\sqrt{2}[/TEX]=sinx 1/ [TEX]\sqrt{2}[/TEX]+cosx 1/[TEX]\sqrt{2}[/TEX]
\Leftrightarrow cos3x= sinx.sin\prod_{i=1}^{n}/4+cosx.cos\prod_{i=1}^{n}/4
\Leftrightarrow cos3x= cos(x-\prod_{i=1}^{n}/4)
do bay h thj co the tu giaj dc uj chu ban :D
 
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