sin a(2cos a-sin a)+sin ^4 a+3cos ^4 a tan a
=2sin acos a-sin^2 a+sin ^4 a+3cos ^3 a sina
=2sin acos a+sin^2 a(1-sin ^2 a)+3cos ^3 a sina
=2sin acos a+sin^2 acos ^2 a+3cos ^3 a sina
=sinacosa(2+sin acos a+3cos^2a)
沒法化了試試同時(shí)除以sin^2 a+cos^2 a？ 答案是這樣解得，但是沒看懂暈，你的題目抄的有問題，我看錯(cuò)題目了你的題目是tana=4，求sin a(2cos a-sin a)+sin ^4 a+3cos ^4 a對(duì)吧？sin a(2cos a-sin a)+sin ^4 a+3cos ^4 a=2sin acos a-sin^2 a+sin ^4 a=2sin acos a+sin^2 a(1-sin ^2 a)=2sin acos a+sin^2 acos ^2 a=2sin acos a+sin^2 acos ^2 a+1-1=(sinacosa+1)^2-1=(sinacosa/1+1)^2-1=[sinacosa/(sin^2a+cos^2a)+1]^2-1=[tana/(1+tan^2a)+1]^2-1=(4/17+1)^2-1=152/289這個(gè).......3cos ^4 asin a(2cos a-sin a)+sin ^4 a+3cos ^4 a=(2sin acos a-sin^2 a)/1+(sin ^4 a+3cos ^4 a)/1=(2sin acos a-sin^2 a)/(sin^2a+cos^2a)+(sin ^4 a+3cos ^4 a)/(sin^2a+cos^2a)=(2tana-tan^2a)/(1+tan^2a)+(sin^2atan^2a+3cos ^2 a)/(1+tan^2a)=(16sin^2a+3cos ^2 a)/17=(16sin^2a+3cos ^2 a)/[17(sin^2a+cos^2a)]=(16tan^2a+3)/[17(tan^2a+1)]=259/289