cos A + cos B + cos C where A B and C are the angles of any triangle
Kehna kia chate ho ?
do we have options/choices for this ?
iwant to simplify it...
\[\frac{ 1 }{ \sec A }+\frac{ 1 }{ \sec B }+\frac{ 1 }{ \sec C }\]
\(\large = 1 + 4\sin{\frac{A}{2}}\sin\frac{B}{2}\sin\frac{C}{2}\)
cos A + cos B + cos C is in the most simplified form
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|
Join our real-time social learning platform and learn together with your friends!