A problem in Trigonometry (Properties of Triangles) v3

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The discussion revolves around proving the equation a^2 b^2 c^2 (sin 2A + sin 2B + sin 2C) = 32 Δ^3 for triangle ABC, where Δ represents the area. The original poster successfully solved the problem with assistance from another website, leading to the thread being marked as solved. However, there is contention regarding the practice of cross-posting on multiple forums, which some users argue wastes others' time and detracts from community support. Despite the criticism, the original poster defends their approach, citing previous benefits from cross-posting. Ultimately, the thread was locked due to the resolution of the problem.
Wrichik Basu
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Homework Statement



In any triangle ABC, prove that $$a^2 b^2 c^2 \left (\sin {2A} +\sin {2B} + \sin {2C} \right) = 32 \Delta ^3$$

Here ##\Delta ## means the area of the triangle.

Homework Equations



The Attempt at a Solution



20170512_113927.jpg
 
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Wrichik Basu said:

Homework Statement



In any triangle ABC, prove that $$a^2 b^2 c^2 \left (\sin {2A} +\sin {2B} + \sin {2C} \right) = 32 \Delta ^3$$

Here ##\Delta ## means the area of the triangle.

Homework Equations



The Attempt at a Solution



View attachment 203411

This thread is marked solved. why ? did you worked this out or you still need help ?
 
As a physicist I notice your change in dimension somewhere halfway (I can point it out if you type it, not if you post a picture...) from length6 to length3 ...
 
Buffu said:
Then it is bad of you to cross post on multiple sites.

Of course I'll cross post, because previously by doing this, several times I've got several different correct ways to solve a single problem, which is interesting.
 
Wrichik Basu said:
Of course I'll cross post, because previously by doing this, several times I've got several different correct ways to solve a single problem, which is interesting.

You don't realize you will waste time of people who will look at this post thinking you need help.
 
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Thread locked for moderation.
 
Wrichik Basu said:
Of course I'll cross post, because previously by doing this, several times I've got several different correct ways to solve a single problem, which is interesting.

We can't stop you from posting on multiple forums, but unfortunately you've probably just lost the help of any of our regular homework helpers that happen to stumble across this thread or that hear about this. As Buffu said, you're mostly just wasting peoples' time.

Thread will remain locked as the problem is, obviously, solved.
 

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