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mathslover
 3 years ago
cos A + cos B + cos C where A B and C are the angles of any triangle
mathslover
 3 years ago
cos A + cos B + cos C where A B and C are the angles of any triangle

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hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1do we have options/choices for this ?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0iwant to simplify it...

hba
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ \sec A }+\frac{ 1 }{ \sec B }+\frac{ 1 }{ \sec C }\]

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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1cos A + cos B + cos C is in the most simplified form

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1in 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
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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
 3 years ago
Best ResponseYou've already chosen the best response.0I 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 : \[\large{\frac{4s^3}{abc}}\]

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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0let me show you my complete work wait..

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1that doesn't hold for an equilateral triangle, all sides equal, say =1

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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1ok, show your work, we'll find the error...

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Oh! I think I got my mistake ... ..

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

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

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0And mine was numerical but only for a right triangle =/

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0=/ yeah @hartnn any try?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1if i get something, i'll post.

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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks! I am also trying my best...

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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0We have to prove for less than or equal to 2

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2why not ?where u have doubt ?

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

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

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2this is because maximum value of cos is 1

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

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2and note i wrote c as 180(A+B)

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2@mathslover you get it?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Are you taking A = B = 0 degrees?

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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0I may be wrong but please correct me if I am ...

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2yep for maximum value we take A=B=C=0 degree

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0but A , B and C are the angles of a triangle..

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2yess that was in general case

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2in the triangle case angle c is replaced by 180(A+B)

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2hence cos C becomes cos (180(A+B))

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Yes but we can not take the angles of a triangle as 0 degrees..

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2i didnt told you to take !!i said that is general case!

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2but equal to not possible

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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0so can cos A + cos B +cos C be greater than 2 also ?

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2no ! thats where we apply trigonometry ! os(180x)=sin x

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2you understand that much?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0cos ( 180  x ) =  cos (x)

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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0got it now thanks a lot

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2:) you are welcome bro!

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1directly 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

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2simple 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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1ohh..in 1st and 4th Quadrant cos is positive....

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2ya whatever value cos takes it wont exceed +2

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

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2its all about the quadrants :)
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