Question

Доказать тождество
$sin^{2}( alpha -2 beta ) - cos^{2} alpha - cos^{2} 2 beta =-2cos alpha *cos2 beta*cos( alpha -2 beta )$

Answer 1

Используем формулы:
cosa*cosb=1/2*(cos(a-b)+cos(a+b))
cos2a=2cos²a-1
sin²a+cos²a=1
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-2cosa*cos2b*cos(a-2b)=-2*1/2(cos(a-2b)+cos(a+2b))*cos(a-2b)=
=-cos²(a-2b)-cos(a+2b)*cos(a-2b)=
=sin²(a-2b)-1-1/2*(cos(a+2b-a+2b)+cos(a-2b+a-2b))=
=sin²(a-2b)-1-1/2cos4b-1/2cos2a=
=sin²(a-2b)-1-1/2*(2cos²2b-1)-1/2*(2cos²a-1)=
=sin²(a-2b)-1-cos²2b+1/2-cos²a+1/2=sin²(a-2b)-cos²2b-cos²a

sin²(a-2b)-cos²a-cos²2b=sin²(a-2b)-cos²2b-cos²a