Question

sin^4a+cos^4a-sin^6a-cos^6a=sin^2a cos^2a докажите тождество

Answer 1

$sin^4(a)+cos^4(a)-sin^6(a)-cos^6(a)=sin^2(a)cos^2(a)\(1-sin^2(a))sin^4(a)+cos^4(a)-cos^6(a)=sin^2(a)cos^2(a)\cos^2(a)+sin^2(a)=1-- textgreater 1-sin^2(a)=cos^2(a)\cos^2(a)sin^4(a)+cos^4(a)-cos^6(a)=sin^2(a)cos^2(a)\(sin^4(a)+cos^2(a)-cos^4(a))cos^2(a)=sin^2(a)cos^2(a)\a^2-b^2=(a-b)(a+b)\((sin^2(a)-cos^2(a))(sin^2(a)+cos^2(a))+cos^2(a))cos^2(a)=\=sin^2(a)cos^2(a)\(-(cos^2(a)-sin^2(a))+cos^2(a))cos^2(a)=sin^2(a)cos^2(a)\(-cos(2a)+cos^2(a))cos^2(a)=sin^2(a)cos^2(a)\cos(2a)=cos^2(a)-sin^2(a)$
$(-(cos^2(a)-sin^2(a))+cos^2(a))cos^2(a)=sin^2(a)cos^2(a)\\(-cos^2(a)+sin^2(a)+cos^2(a))cos^2(a)=sin^2(a)cos^2(a)\sin^2(a)cos^2(a)=sin^2(a)cos^2(a)$