Question

докажите тождество
1)(2-sina)(2+sina)+(2-cosa)(2+cosa)=7
2)sin^4a-cos^4a=sin^2a-cos^2a.

Answer 1

$(2-sinalpha)(2+sinalpha)+(2-cosalpha)(2+cosalpha)=7\2^2-sin^2alpha+2^2-cos^2alpha=7\sin^2alpha+cos^2alpha=1\\sin^4alpha-cos^4alpha=sin^2alpha-cos^2alpha\sin^4alpha-sin^2alpha-cos^4alpha+cos^2alpha=0\sin^2alpha(sin^2alpha-1)-cos^2alpha(cos^2alpha-1)=0\sin^2alpha*(-cos^2alpha)-cos^2alpha*(-sin^2alpha)=0\-sin^2alpha cos^2alpha+sin^2alpha cos^2alpha=0$