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

Mathematics
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  • hba
Kehna kia chate ho ?
do we have options/choices for this ?
iwant to simplify it...

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Other answers:

  • hba
\[\frac{ 1 }{ \sec A }+\frac{ 1 }{ \sec B }+\frac{ 1 }{ \sec C }\]
\(\large = 1 + 4\sin{\frac{A}{2}}\sin\frac{B}{2}\sin\frac{C}{2}\)
cos A + cos B + cos C is in the most simplified form
  • hba
Yeah right :)
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.
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
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 : \[\large{\frac{-4s^3}{abc}}\]
you mean u got \(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2-a^3-b^3-c^3=-8s^3\)
let me show you my complete work wait..
that doesn't hold for an equilateral triangle, all sides equal, say =1
Yes but I am getting this answer by simplifying cos A + cos B + cos C by using cosines law..
ok, show your work, we'll find the error...
ok 1 minute
Oh! I think I got my mistake ... ..
can you help me to prove that cos A + cos B + cos C is less than or equal to 2
I got some conceptual answers previous time but I am searching for a numerical one.. @hartnn
And mine was numerical but only for a right triangle =/
=/ yeah @hartnn any try?
if i get something, i'll post.
\[{cos A + cos B + cos C = \\ \frac{a(b^2+c^2) + b(a^2+c^2) + \\c(a^2+b^2) - (a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{2abc}}\]
trying on paper
Thanks! I am also trying my best...
got it !
?
|dw:1356616135337:dw|
yes..
now cos A+cos B+Cos C is always less than equal to 3
We have to prove for less than or equal to 2
ya i am proving
ok
|dw:1356616236014:dw|
you get that >?
No :(
why not ?where u have doubt ?
cos A + cos B - cos(A+B) , how is it less than or equal to 3 ?
i wrote cos A+cos B+cos(18-(A+B)<=3
this is because maximum value of cos is 1
ie 1+1+1=3
all other cases cos a +cos b +cos c value will be less than 3
and note i wrote c as 180-(A+B)
@mathslover you get it?
wait!
ok take your time !
Are you taking A = B = 0 degrees?
since cos A + cos B + cos (180-(A+B)) = 3 is not possible ?
I may be wrong but please correct me if I am ...
yep for maximum value we take A=B=C=0 degree
but A , B and C are the angles of a triangle..
yess that was in general case
in the triangle case angle c is replaced by 180-(A+B)
hence cos C becomes cos (180-(A+B))
Yes but we can not take the angles of a triangle as 0 degrees..
i didnt told you to take !!i said that is general case!
but equal to not possible
in triangle you have equation cos A+cos B+cos(180-(A+B))<3
so can cos A + cos B +cos C be greater than 2 also ?
no ! thats where we apply trigonometry ! os(180-x)=-sin x
you understand that much?
cos ( 180 - x ) = - cos (x)
oops srry ya
-cos x
so our equation for triangle becomes cos A+cos B-cos(A+B)<3
got it now thanks a lot
:) you are welcome bro!
directly comes from,\(\cos(A+B)=-\cos C\) than from cos A+cos B-cos(A+B)<3 how u proved cos A+cos B+cos C<=2
?
simple cos A +cos B maximum can be 2 and -cos(a+b) is a reducing term so it always results in value less than or equal to 2 for the expression
cos can be negative
ohh..in 1st and 4th Quadrant cos is positive....
ya whatever value cos takes it wont exceed +2
yep !
i thought cos(A+B) could be negative, thats why i doubted..but cos (A+B) will be positive...
;)
its all about the quadrants :)

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