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Mathematics 16 Online
mathslover (mathslover):

cos A + cos B + cos C where A B and C are the angles of any triangle

OpenStudy (hba):

Kehna kia chate ho ?

hartnn (hartnn):

do we have options/choices for this ?

mathslover (mathslover):

iwant to simplify it...

OpenStudy (hba):

\[\frac{ 1 }{ \sec A }+\frac{ 1 }{ \sec B }+\frac{ 1 }{ \sec C }\]

OpenStudy (foolaroundmath):

\(\large = 1 + 4\sin{\frac{A}{2}}\sin\frac{B}{2}\sin\frac{C}{2}\)

hartnn (hartnn):

cos A + cos B + cos C is in the most simplified form

OpenStudy (hba):

Yeah right :)

hartnn (hartnn):

in any triangle , A+B+C = \(\pi\), so \(\cos(A+B)=-\cos C\) \(\cos(C+B)=-\cos A\) \(\cos(A+C)=-\cos B\) if you want to complicate, i can help :P using cosine law.

mathslover (mathslover):

Well yes I wanted to use cosine law here and I got : cos A + cos B + cos C = \(\LARGE{-\frac{4s^3}{abc}}\) where a , b and c are the sides of the triangle and s is semi perimeter of the triangle

mathslover (mathslover):

I did like this : \[\large{\frac{(b^2+c^2-a^2)}{2bc} + \frac{(c^2+a^2-b^2)}{2ac} + \frac{(a^2+b^2-c^2)}{2ab}}\] and simpified this by taking LCM and taking terms common and I got : \[\large{\frac{-4s^3}{abc}}\]

hartnn (hartnn):

you mean u got \(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2-a^3-b^3-c^3=-8s^3\)

mathslover (mathslover):

let me show you my complete work wait..

hartnn (hartnn):

that doesn't hold for an equilateral triangle, all sides equal, say =1

mathslover (mathslover):

Yes but I am getting this answer by simplifying cos A + cos B + cos C by using cosines law..

hartnn (hartnn):

ok, show your work, we'll find the error...

mathslover (mathslover):

ok 1 minute

mathslover (mathslover):

Oh! I think I got my mistake ... ..

mathslover (mathslover):

can you help me to prove that cos A + cos B + cos C is less than or equal to 2

mathslover (mathslover):

I got some conceptual answers previous time but I am searching for a numerical one.. @hartnn

Parth (parthkohli):

And mine was numerical but only for a right triangle =/

mathslover (mathslover):

=/ yeah @hartnn any try?

hartnn (hartnn):

if i get something, i'll post.

mathslover (mathslover):

\[{cos A + cos B + cos C = \\ \frac{a(b^2+c^2) + b(a^2+c^2) + \\c(a^2+b^2) - (a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{2abc}}\]

OpenStudy (aravindg):

trying on paper

mathslover (mathslover):

Thanks! I am also trying my best...

OpenStudy (aravindg):

got it !

mathslover (mathslover):

?

OpenStudy (aravindg):

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