blocks|key|834587|text|我们有acos+function，它返回以弧度表示的角度。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|834588|>>>+import+math
>>>+math.acos(0)
1.5707963267948966
>>>+_+*+2+-+math.pi
0.0|code-block|syntax|javascript|834589|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/math.html#math.acos^0|3|4|3|D|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

We have the <a href="http://docs.python.org/library/math.html#math.acos" rel="noreferrer"><code>acos</code> function</a>, which returns the angle in radians.

<pre><code>&gt;&gt;&gt; import math
&gt;&gt;&gt; math.acos(0)
1.5707963267948966
&gt;&gt;&gt; _ * 2 - math.pi
0.0
</code></pre>

blocks|key|619064|text|您正在寻找math.acos()函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|619065|entityMap^0|5|B|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

You're looking for the <code>math.acos()</code> function.

blocks|key|619098|text|或者只需为cos%5E{-1}的taylor+expansion编写自己的函数|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|619099|这可能会更耗时(运行速度也可能更慢)，但这是更通用的方法|619100|entityMap|0|LINK|mutability|MUTABLE|url|http://mathworld.wolfram.com/InverseCosine.html^0|E|G|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]]|C|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]]]

or just write a function of your own for the <a href="http://mathworld.wolfram.com/InverseCosine.html" rel="nofollow">taylor expansion</a> of cos^{-1}

this would be more time consuming (and maybe slower to run) but is the more general approach

blocks|key|619150|text|要增加使用math.acos的正确答案，还需要知道cmath中有适用于复数的数学函数|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|619151|>>>+import+cmath
>>>+cmath.acos(1j)
(1.5707963267948966-0.88137358701954294j)|code-block|syntax|javascript|619152|如果您只对实数感兴趣，请坚持使用math.acos。|619153|entityMap^0|5|9|P|5|0|0|G|9|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

To augment the correct answers to use <code>math.acos</code>, it is also worth knowing that there are math functions suitable for complex numbers in <code>cmath</code>:

<pre><code>&gt;&gt;&gt; import cmath
&gt;&gt;&gt; cmath.acos(1j)
(1.5707963267948966-0.88137358701954294j)
</code></pre>

Stick with <code>math.acos</code> if you're only interested in real numbers,

blocks|key|834773|text|为了响应使用反余弦通过math.acos找到返回角度，只要角度是<=90*，这是很好的。一旦超过那个角度，python将无法区分您想要的角度。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|834774|观察。|834775|>>>+math.cos(5)
0.28366218546322625|code-block|syntax|javascript|834776|上面，我让python给我取一个5弧度角的余弦，它给了我一个很好的，下面我会让python给我一个有.28~余弦的弧度。应该是5，对吧？它真的告诉我这是真的。|834777|>>>+math.acos(0.28366218546322625)
1.2831853071795865|834778|不对!Python返回1.28~弧度。原因很明显，当可视化绘图时，1.28rad与5rad具有相同的余弦，它们是相反的角度。每个角度与另一个角度共享相同的正弦(与另外两个角度共享-sine+)。即5/175*共用一个等价的正弦。它们的余弦成反比，分别为0.99~-0.99。他们的-sine近亲将是185和355。这里的关系模因是，所有这些角度都与水平轴有相同的角度偏转。5*。|834779|python返回1.28而不是5的原因是所有的计算机/计算器都是基于一个类似算盘的数据表，包括角度/弧度、它的正弦、cos、tan等。所以当I+math.acos(x)时，python要求内核在数据表中查找哪个角度有一个余弦x，当它找到它时，它会返回它出现的第一个条目。然后python给了我这个角度。|834780|由于这种共享的、成比例的对称性，sin/cos比率经常重复。而且你很可能会多次看到相同的图形。如果不执行额外的逻辑来考虑角度的正弦值-/%2B，python或操作系统就无法确定您实际需要的两个角度中的哪个角度的差异。或者角度是相切的。|834781|1.28+Rad+has++x+cosine,+y+sine,+z+tan++(72*)
1.88+Rad+has+-x+cosine,+y+sine,+-z+tan+(108*)
4.39+Rad+has+-x+cosine,+-y+sine,+z+tan+(252*)
+++5+Rad+has++x+cosine,+-y+sine,+-z+tan+(288*)|834782|或者，从笛卡尔的角度来看，|834783|+++++++++++++++++++++++negX,posY+%7C+posX,posY
++++++++++++++++++++++++++++-----%2B-----
+++++++++++++++++++++++negX,negY+%7C++posX,negY

1.88+Rad+has+-x+cosine,+y+sine+(108)+%7C+1.28+Rad+has++x+cosine,+y+sine+(72*)
++++++++++++++++++++++++++++++++-----%2B-----
4.39+Rad+has+-x+cosine,+-y+sine+(252)%7C++++5+Rad+has++x+cosine,+-y+sine+(288)|834784|因此，无论出于什么原因，如果我需要选择5个弧度(例如，对于矢量绘图或游戏，以确定敌人来自玩家的各种矢量)，我将不得不进行某种类型的if/then逻辑比较正弦/正切。|834785|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|15|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|19|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|1A|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1B|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1C|8|@]|9|@]|A|$G|H]]|$1|Y|3|Z|5|6|7|1D|8|@]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|11|$]]

In response to using inverse cosine to find return angles via math.acos, it's all fine and dandy so long as the angle is &lt;=90* once you go past that, python will have no way of differentiating which angle you wanted.

Observe. 

<pre><code>&gt;&gt;&gt; math.cos(5)
0.28366218546322625
</code></pre>

Above, I asked python to fetch me the cosine of a 5 radian angle, and it gave me .28~ Great, below I'll ask python to give me the radian which has a .28~ cosine. Should be 5, right? It literally just told me it was.

<pre><code>&gt;&gt;&gt; math.acos(0.28366218546322625)
1.2831853071795865
</code></pre>

Wrong! Python returns 1.28~ radians. The reason is obvious when plotted visually, 1.28rad has the same cosine as 5rad, they're inverse angles. Every angle shares the same sine with another angle (and -sine with two others).
i.e. 5/175* share an equivalent sine. They share inversely proportional cosines .99~/-.99 respectively. Their -sine cousins would be 185 and 355. The relationship meme here is that all these angles share the same angular deflection from the horizontal axis. 5*.

The reason python returns 1.28 and not 5 is that all computers/calculators are based on an abacus-like data table of an angle/radian, its sine, cos, tan etc etc. So when i math.acos(x), python asks the kernal to look through that data table for whichever angle has a cosine of x, and when it finds it, it returns the first entry it appears with. and then python gives that angle to me.

Due to this shared, proportional symmetry, sin/cos ratios repeat frequently. And you are likely to see the same figure, multiple times. There's no way for python, or the OS, to determine the difference of which of the two angles you actually require without doing additional logic that takes into account the -/+ value of the angle's sine. Or, the angle's tangent.

<pre><code>1.28 Rad has x cosine, y sine, z tan (72*)
1.88 Rad has -x cosine, y sine, -z tan (108*)
4.39 Rad has -x cosine, -y sine, z tan (252*)
 5 Rad has x cosine, -y sine, -z tan (288*)
</code></pre>

or, viewed Cartesianly,

<pre><code> negX,posY | posX,posY
 -----+-----
 negX,negY | posX,negY

1.88 Rad has -x cosine, y sine (108) | 1.28 Rad has x cosine, y sine (72*)
 -----+-----
4.39 Rad has -x cosine, -y sine (252)| 5 Rad has x cosine, -y sine (288)
</code></pre>

So if, for whatever reason, i need 5 radians to be chosen (say for a vector drawing or game to determine the various vectors enemies are from the player), i would have to do some type of if/then logic comparing the sines/tangents.

blocks|key|619287|text|math.acos()的结果是弧度。所以你需要把它转换成度数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|619288|以下是您可以执行的操作：|619289|import+math
res+=+math.degrees+(math.acos+(value))|code-block|syntax|javascript|619290|entityMap^0|0|B|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|S|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

The result of <code>math.acos()</code> is in radians. So you need to convert that to degrees.

Here is how you can do:

<pre><code>import math
res = math.degrees (math.acos (value))
</code></pre>

Apologies if this is straight forward, but I have not found any help in the python manual or google.
I am trying to find the inverse cosine for a value using python.
i.e. cos⁻¹(x)
Does anyone know how to do this?
Thanks

Inverse Cosine in Python

如果这是直截了当的，我很抱歉，但我在python手册或google中没有找到任何帮助。我正在尝试用python找一个值的反余弦。即cos⁻²(X)有人知道怎么做吗？谢谢

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