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mathslover
Group Title
cos A + cos B + cos C where A B and C are the angles of any triangle
 one year ago
 one year ago
mathslover Group Title
cos A + cos B + cos C where A B and C are the angles of any triangle
 one year ago
 one year ago

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hba Group TitleBest ResponseYou've already chosen the best response.0
Kehna kia chate ho ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
do we have options/choices for this ?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
iwant to simplify it...
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ 1 }{ \sec A }+\frac{ 1 }{ \sec B }+\frac{ 1 }{ \sec C }\]
 one year ago

FoolAroundMath Group TitleBest ResponseYou've already chosen the best response.0
\(\large = 1 + 4\sin{\frac{A}{2}}\sin\frac{B}{2}\sin\frac{C}{2}\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
cos A + cos B + cos C is in the most simplified form
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
I did like this : \[\large{\frac{(b^2+c^2a^2)}{2bc} + \frac{(c^2+a^2b^2)}{2ac} + \frac{(a^2+b^2c^2)}{2ab}}\] and simpified this by taking LCM and taking terms common and I got : \[\large{\frac{4s^3}{abc}}\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
you mean u got \(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2a^3b^3c^3=8s^3\)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
let me show you my complete work wait..
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
that doesn't hold for an equilateral triangle, all sides equal, say =1
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Yes but I am getting this answer by simplifying cos A + cos B + cos C by using cosines law..
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
ok, show your work, we'll find the error...
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
ok 1 minute
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Oh! I think I got my mistake ... ..
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
can you help me to prove that cos A + cos B + cos C is less than or equal to 2
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
I got some conceptual answers previous time but I am searching for a numerical one.. @hartnn
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
And mine was numerical but only for a right triangle =/
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
=/ yeah @hartnn any try?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
if i get something, i'll post.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
\[{cos A + cos B + cos C = \\ \frac{a(b^2+c^2) + b(a^2+c^2) + \\c(a^2+b^2)  (a+b+c)(a^2+b^2+c^2abbcca)}{2abc}}\]
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
trying on paper
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Thanks! I am also trying my best...
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
dw:1356616135337:dw
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
now cos A+cos B+Cos C is always less than equal to 3
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
We have to prove for less than or equal to 2
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
ya i am proving
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
dw:1356616236014:dw
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
you get that >?
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
why not ?where u have doubt ?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
cos A + cos B  cos(A+B) , how is it less than or equal to 3 ?
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
i wrote cos A+cos B+cos(18(A+B)<=3
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
this is because maximum value of cos is 1
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
ie 1+1+1=3
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
all other cases cos a +cos b +cos c value will be less than 3
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
and note i wrote c as 180(A+B)
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
@mathslover you get it?
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
ok take your time !
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Are you taking A = B = 0 degrees?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
since cos A + cos B + cos (180(A+B)) = 3 is not possible ?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
I may be wrong but please correct me if I am ...
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
yep for maximum value we take A=B=C=0 degree
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
but A , B and C are the angles of a triangle..
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
yess that was in general case
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
in the triangle case angle c is replaced by 180(A+B)
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
hence cos C becomes cos (180(A+B))
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Yes but we can not take the angles of a triangle as 0 degrees..
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
i didnt told you to take !!i said that is general case!
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
but equal to not possible
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
in triangle you have equation cos A+cos B+cos(180(A+B))<3
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
so can cos A + cos B +cos C be greater than 2 also ?
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
no ! thats where we apply trigonometry ! os(180x)=sin x
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
you understand that much?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
cos ( 180  x ) =  cos (x)
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
oops srry ya
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
so our equation for triangle becomes cos A+cos Bcos(A+B)<3
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
got it now thanks a lot
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
:) you are welcome bro!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
directly comes from,\(\cos(A+B)=\cos C\) than from cos A+cos Bcos(A+B)<3 how u proved cos A+cos B+cos C<=2
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
cos can be negative
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
ohh..in 1st and 4th Quadrant cos is positive....
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
ya whatever value cos takes it wont exceed +2
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
i thought cos(A+B) could be negative, thats why i doubted..but cos (A+B) will be positive...
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.2
its all about the quadrants :)
 one year ago
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