必修四的三角涵数帮我求证：1-2sinxcosx/cos^2-sin^2=1-tanx/1+tanxsin^4x+cos^4x=1-2sin^2xcos^2x谢谢,最好给我提示让我自己做

Question

必修四的三角涵数
帮我求证：1-2sinxcosx/cos^2-sin^2=1-tanx/1+tanx
sin^4x+cos^4x=1-2sin^2xcos^2x
谢谢,最好给我提示让我自己做

Answer 1

1.是(1-2sinxcosx)/(cos^2x-sin2^x)=(1-tanx)/(1+tanx)吧!
左分子(1-2sinxcosx)=(cos^x-sin^x)^2
左分母(cos^2x-sin2^x)=(cos^x-sin^x)(cos^x+sin^x)
则左=(cos^x-sin^x)/(cos^x+sin^x)
右边切割化弦也得(cos^x-sin^x)/(cos^x+sin^x)
故:(1-2sinxcosx)/(cos^2x-sin2^x)=(1-tanx)/(1+tanx)
2.
1-2sin^2xcos^2x=(sin^2x+cos^2x)^2-2sin^2xcos^2x
=sin^4x+cos^4x