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 January 11th, 2019, 06:56 PM #11 Global Moderator   Joined: Dec 2006 Posts: 21,127 Thanks: 2336 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. Thanks from topsquark January 13th, 2019, 10:52 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, 02:15 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, 02:17 PM   #14
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 Originally Posted by skipjack 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.
Thanks January 13th, 2019, 07:50 PM #15 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Regarde dans le dictionnaire Larousse  January 13th, 2019, 10:52 PM   #16
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 Originally Posted by Larrousse Correction for the fourth line. $(b\times\cos A)^2+(a\times\cos B)^2=c^2$
That's incorrect. I'm not sure what you intended. January 14th, 2019, 01:52 AM   #17
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 Originally Posted by skipjack That's incorrect. I'm not sure what you intended.
It's an error, disregard it. Tags detailed, magical, properties, section, triangles, unexplained Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Carl James Mesaros Computer Science 2 April 13th, 2017 01:49 AM BenFRayfield Computer Science 0 February 11th, 2015 09:06 PM qskti Linear Algebra 0 January 9th, 2015 10:57 AM jonas Algebra 1 November 12th, 2009 11:55 PM

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