1+sin4θ−cos4θ |
2tanθ |
1+sin4θ+cos4θ |
1−tan2θ |
只需证
1+2sin2θcos2θ−(1−2sin22θ) |
2tanθ |
1+2sin2θcos2θ+2cos22θ−1 |
1−tan2θ |
即证
2sin2θ(sin2θ+cos2θ) |
2tanθ |
2cos2θ(sin2θ+cos2θ) |
1−tan2θ |
即证
sin2θ |
2tanθ |
cos2θ |
1−tan2θ |
sin2θ |
cos2θ |
2tanθ |
1−tan2θ |
只需证tan2θ=
2tanθ |
1−tan2θ |
由二倍角的正切公式可知上式正确,
故原命题得证.