Now I give my proof.

notice that $$$\sin(x)\sin(y)=\frac{1}{2}(cos(x-y)-cos(x+y))$$$

inverse $$$\sin(x)\sin(y)=\frac{1}{2}(cos(x-y)-cos(x+y))$$$ to be $$$\cos(a)-\cos(b)=-2\sin(\frac{a+b}2)\sin(\frac{a-b}2)$$$

maybe a nice idea! What do u think?

# | User | Rating |
---|---|---|

1 | tourist | 3851 |

2 | jiangly | 3634 |

3 | Um_nik | 3539 |

4 | slime | 3498 |

5 | ksun48 | 3493 |

6 | djq_cpp | 3486 |

7 | maroonrk | 3471 |

8 | MiracleFaFa | 3466 |

9 | Radewoosh | 3442 |

10 | Petr | 3426 |

# | User | Contrib. |
---|---|---|

1 | -is-this-fft- | 183 |

2 | awoo | 181 |

3 | YouKn0wWho | 177 |

4 | Um_nik | 175 |

5 | dario2994 | 172 |

6 | Monogon | 170 |

6 | adamant | 170 |

8 | maroonrk | 169 |

9 | errorgorn | 166 |

10 | antontrygubO_o | 165 |

Now I give my proof.

notice that $$$\sin(x)\sin(y)=\frac{1}{2}(cos(x-y)-cos(x+y))$$$

inverse $$$\sin(x)\sin(y)=\frac{1}{2}(cos(x-y)-cos(x+y))$$$ to be $$$\cos(a)-\cos(b)=-2\sin(\frac{a+b}2)\sin(\frac{a-b}2)$$$

maybe a nice idea! What do u think?

↑

↓

Codeforces (c) Copyright 2010-2022 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Aug/18/2022 23:16:19 (g2).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|

Isn't it taught at high school?

Sometimes I am grateful, that I don't study in such schools.

I heard chineseman learn something about it in primary, craaaaaaaazy

As a Chinese, I can tell you that most of us don't. (a huuuge Orz to any primary school students kicking my butt at math though)

you dont know, Botswana knows

this is liexiang, do you kno liexiang

plz support Botswana

I do know about 裂项(liexiang), and although in China primary school students do learn about it, it's restricted to fractional kinds, but this one is trig, and we don't even know about the existence of trig until 9th grade...

裂项(lièxiàng) in Chinese is translated to telescoping.

See also: Wikipedia and Section 2.6 in Concrete Mathematics. (IDK why in one Chinese version it was translated into 叠缩)

That's too exaggerated，We Just Learn this at junior high school..

LS Dini Pisa orz

Snob

LS Dini Pisa orz

not most of the times sir...

Random Chinese guy comes and say: 'Well I learnt it when I was six, lol noob' (not me though)

btw this is a pretty neat trig conclusion.

Yes it's a nice identity though it's fairly standard. You can also use complex numbers to get the same result with $$$\sum_{k=1}^{n} \sin(kx) = \text{Im} \sum_{k=1}^{n} e^{i k \theta}$$$

wow! A even nicer proof, clearer than me!

It's a very well-known thing in optics when you have rays of light interfering, just with real component (sum of cosines) rather than imaginary.

Another trigonometric sum I found mildly interesting is,

Which comes from, $$$\cot{x} - 2 \cot 2x = \tan x$$$

Don't care + didn't ask

who cares about your opinion?

Well, the author asked "What do u think?", so maybe he does care. My point was that trig is awful and kind of not related to cp. This is, of course, very subjective.

yeah but the author didn't ask if you care or no, and he also didn't ask if u asked.

do u even understand what does "What do u think?" even mean lol?

ratio

Completely unrelated response:

SpoilerAlbedo is love

12th class stuff!!

For us Asians.

If you sum up $$$\cos kx$$$ rather than $$$\sin kx$$$, you'll end up with a family of functions known as the Dirichlet kernel.

Dirichlet kernel has a great importance in the Fourier analysis, as the convolution of any function with $$$n$$$-th Dirichlet kernel will provide the $$$n$$$-th degree Fourier approximation of the function.

I would personally prefer the complex numbers way to compute the sum:

Multiplying the numerator and the denominator by $$$e^{-\frac{ix}{2}}$$$ and using $$$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$$$ formula, we get

The real part of the nominator is

The imaginary part of the nominator is

Therefore,

and