直接泰勒展开（滑稽）
by zhanghengrui @ 2019-03-23 23:21:34
~~去你的萌新~~
by zhanghengrui @ 2019-03-23 23:21:58
有的。
$$\large G(x)=\tan F(x)$$
$$\large G'(x)=\frac{F'(x)}{\cos^2F(x)}$$
$$\large G'(x)=\int\frac{F'(x)}{\cos^2F(x)}dx$$
这样只用算一遍三角函数
by NaCly_Fish @ 2019-03-23 23:25:45
@[NaCly_Fish](/space/show?uid=115864) 最后一个式子打错了。
应该是
$$\large G(x)=\int\frac{F'(x)}{\cos^2F(x)}dx$$
@[SLF_LLL_SPFA](/space/show?uid=119553)
by NaCly_Fish @ 2019-03-23 23:26:27
@[SLF_LLL_SPFA](/space/show?uid=119553) 去你妈的萌新
by yzhang @ 2019-03-23 23:27:47
至于有没有更快的方法，窝就不清楚了qaq
by NaCly_Fish @ 2019-03-23 23:28:54
@[SLF_LLL_SPFA](/space/show?uid=119553)
$$tan\space\alpha=cot\space \alpha$$
滑稽
by Jelly_Goat @ 2019-03-23 23:56:16
不对应该是
$\huge{tan \space\alpha={cot\space \alpha\over 1}}$
by Jelly_Goat @ 2019-03-23 23:57:53
还是不对
$\huge{tan\space\alpha={1\over cot\space\alpha} }$
~~今天手感不好~~
by Jelly_Goat @ 2019-03-23 23:59:19
@[NaCly_Fish](/space/show?uid=115864) zz了。。。然后代码爆炸啊。。
by rEdWhitE_uMbrElla @ 2019-03-24 00:08:48