A beautiful equation

Indeed the saying "Curiosity is the mother of inventions" is absolutely true. While learning trigonometry , I found this:

Prove that in ΔABC\Delta ABC ,

cyccotA=cyccscA+cyccotA\large\sum_{cyc} \cot A = \prod_{cyc} \csc A + \prod_{cyc} \cot A

I invite all brilliantians to prove this :)

Note by Nihar Mahajan
3 years, 10 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Since, A+B+C=180oA+B+C = 180^{o}

Cos(A+B+C)=1Cos(A+B+C) = -1

cyccos(A)cyccos(A)sin(B)sin(C)=1 \prod_{cyc} \cos (A) - \sum_{cyc} \cos (A) \sin (B) \sin (C) = -1

Divide both sides with cycsin(A)\prod_{cyc} \sin (A),

cyccot(A)cyccot(A)=cyccsc(A)\prod_{cyc} \cot (A) - \sum_{cyc} \cot (A) = - \prod_{cyc} \csc (A)

cyccot(A)=cyccot(A)+cyccsc(A)\sum_{cyc} \cot (A) = \prod_{cyc} \cot (A) + \prod_{cyc} \csc (A)

Hence, proved.

Surya Prakash - 3 years, 10 months ago

Log in to reply

@Nihar Mahajan How is my solution? Is there any other method u r thinking?

Surya Prakash - 3 years, 10 months ago

Log in to reply

It is quite shorter and elegant than mine.Nice work!

Nihar Mahajan - 3 years, 10 months ago

Log in to reply

@Nihar Mahajan Do u have another method? Can u please post it?

Surya Prakash - 3 years, 10 months ago

Log in to reply

@Surya Prakash I have another method , but it does not use basic results. It uses "lemma type" results , thats because I found this equation while proving that lemma only!

Nihar Mahajan - 3 years, 10 months ago

Log in to reply

@Nihar Mahajan Please send that solution . I am eager to know another solution to it. Please

Surya Prakash - 3 years, 10 months ago

Log in to reply

@Nihar Mahajan Wait isn't that what I had told you?

Sudeep Salgia - 3 years, 10 months ago

Log in to reply

@Sudeep Salgia That day , we both forgot that cos(A+B+C)=1\cos(A+B+C)=-1. Don't worry , it happens sometimes :P

Nihar Mahajan - 3 years, 10 months ago

Log in to reply

Hello Nihar! Can you please tell me from where are you learning trigonometry?? Thanks 😀

Harsh Shrivastava - 3 years, 10 months ago

Log in to reply

M Prakash Academy , Pune.

Nihar Mahajan - 3 years, 10 months ago

Log in to reply

Exactly what is "cyccyc" in cyc\displaystyle \sum_{cyc} and cyc\displaystyle\prod_{cyc}?    ~~~ Never seen this.

Micah Wood - 3 years, 10 months ago

Log in to reply

cyccotA=cotA+cotB+cotCcyccotA=cotAcotBcotC\sum_{cyc} \cot A = \cot A+\cot B+\cot C \\ \prod_{cyc} \cot A = \cot A\cot B\cot C

It is short form of the cyclic summation/product.I hope you get it now :)

Nihar Mahajan - 3 years, 10 months ago

Log in to reply

what does sym summation mean?

Dev Sharma - 3 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...