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prove that [tex]cos 3 x =4 cos ^{3}x-3 cos x[/tex]

Sagot :

[tex]cos3x=4cos^3x-3cosx\\\\L=cos(2x+x)=cos2xcosx-sin2xsinx\\\\=(cos^2x-sin^2x)cosx-2sinxcosxsinx\\\\=cos^3x-sin^2xcosx-2sin^2xcosx\\\\=cos^3x-3sin^2xcosx\\\\=cos^3x-3cosx(1-cos^2x)\\\\=cos^3x-3cosx+3cos^3x\\\\=4cos^3x-3cosx=R[/tex]


[tex]cos(a+b)=cosa\ cosb-sina\ sinb\\\\sin^2a+cos^2b=1\to sin^2a=1-cos^2b\\\\cos2a=cos^2a-sin^2a\\\\sin2a=2sina\ cosa[/tex]
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