January 11th, 2019, 05:56 PM  #11 
Global Moderator Joined: Dec 2006 Posts: 20,757 Thanks: 2138 
The inequalities $a+b>c,\, a+c>b,\, b+c>a$ are known as the triangle inequality. Your third and fourth identities can be written as $(c\times\sin B)^2+(c\times\cos B)^2=c^2$ and $(c\times\sin A)^2+(c\times\cos A)^2=c^2$ so that they hold for any value of $c$, not just when $c$ is 1. Hence my change to this effect in your previous list. As $c\times\sin B = b\times\sin C$, $(c\times\sin B)^2+(b\times\cos C)^2=b^2$ is equivalent to $(b\times\sin C)^2+(b\times\cos C)^2=b^2$, which can be simplified to $\sin^2\! C + \cos^2\! C = 1$. I think all the identities you listed can be found on Wikipedia. 
January 13th, 2019, 09:52 AM  #12 
Newbie Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 
Correction for the fourth line. $(b\times\cos A)^2+(a\times\cos B)^2=c^2$ 
January 13th, 2019, 01:15 PM  #13 
Newbie Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 
For side c so the base is not only 1the equations are: $(c\times\sin B)^2+(c\times\cos B)^2=c^2$ $(c\times\sin A)^2+(c\times\cos A)^2=c^2$ 
January 13th, 2019, 01:17 PM  #14  
Newbie Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0  Quote:
 
January 13th, 2019, 06:50 PM  #15 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
Regarde dans le dictionnaire Larousse 
January 13th, 2019, 09:52 PM  #16 
Global Moderator Joined: Dec 2006 Posts: 20,757 Thanks: 2138  
January 14th, 2019, 12:52 AM  #17 
Newbie Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0  

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