来自蒋纯的问题
sin^2β+cos^4β+sin^2βcos^2β
sin^2β+cos^4β+sin^2βcos^2β
1回答
2020-08-02 13:34
sin^2β+cos^4β+sin^2βcos^2β
sin^2β+cos^4β+sin^2βcos^2β
b=β,sin^2(b)+cos^2(b)=1
sin^2(b)+cos^4(b)+sin^2(b)*cos^2(b)
=sin^2(b)+[cos^4(b)+sin^2(b)*cos^2(b)]
=sin^2(b)+cos^2(b)*[cos^2(b)+sin^2(b)]
=sin^2(b)+cos^2(b)*1
=sin^2(b)+cos^2(b)
=1