Question

помогите пожалуйста,sin^6a-cos^6a-3/4cos^2a+3/4sin^2a

Answer 1

sin^6α-cos^6α-(3/4)*cos²2α+(3/4)*sin²α
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sin^6α-cos^6α-(3/4)*cos²α+(3/4)*sin²α= -(cos^6α-sin^6α)-(3/4)(cos²α-sin²α)=
 -(cos²α-sin²α)(cos^4α+sin²α*cos²α+sin^4α) -3/4(cos²α-sin²α)=
 -(cos²α-sin²α)(cos^4α+sin²α*cos²α+sin^4α -3/4)=
 -cos2α((cos²α+sin²α)²-cos²α*sin²α -3/4)= - cos2α(1-(1/4)*sin²2α -3/4)=
 - cos2α(1/4-(1/4)*sin²2α)= - (1/4)cos2α*(1-sin²2α) = - (1/4)cos2α*cos²2α=
 - (1/4)cos³2α .