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kasenova19
@kasenova19
July 2022
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вычислите:3cos в квадрате альфа-6+3 sin в квадрате альфа;при cos альфа=0,3
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вион
(sin^2a+3cos^2a)^2+(cos^2a-3sin^2a)^2=(1+2cos^2a)^2+(4cos^2a-3)^2==1+4cos^4a+4cos^2a+16cos^4a+9-24cos^2a=20cos^4a-20cos^2a+10==10(2cos^4a-2cos^2a+1)=10*(2cos^4a-cos2a)x/y=2y/x=0,5(4y^2-3xy+x^2)/(x^2-xy+y^2)=y^2[4-3x/y+(x/y)^2]/x^2[1-y/x+(y/x)^2]=
=0,25*[4-6+4]/[1-0,5+0,25]=0,5/0,75=2/3
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