Автор - zamirka95

упростить cos2a+cos4a/sin2a+sin2a

Ответ

Автор - wieslawlozickip5w1nd

Воспользуем формулy

1)sinx+siny=2sin(x+y)/2 *cos(x-y)/2

2)cosx+cosy=2cos(x+y)/2*cos(x-y)/2

              &      

(cos2a+cos4a)/(sin2a+cos4a)=

*(cos2a+cos4a)=2[cos((2a+4a)/2)]*[cos((2a-4a)/2] =2cos3a*cos(-a)

*(sin2a+cos4a)=2[sin(4a+2a)/2]*[cos(2a-4a)/2]=2sin3a*cos(-a)

=2cos3a*cos(-a) / 2sin3a*cos(-a) =

=cos3a/sin3a=    tg3a


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