In △ABC cos A+cos B+cos C=(3/2) prove that trianle is equilateral


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Question Number 54502 by math1967 last updated on 05/Feb/19

$I n △ A B C \cos A + \cos B + \cos C = \frac{3}{2}$ $p r o v e t h a t t r i a n l e i s e q u i l a t e r a l$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19
$point A,B,C lies on cosx curve point A coordinate is (A,cosA) point B coordinate (B,cosB) point C coordinate(C,cosC) △ABC centroid is G coordinate of G is(((x_1 +x_2 +x_3 )/3),((y_1 +y_2 +y_3 )/3)) so G is(((A+B+C)/3),((cosA+cosB+cosC)/3)) as per diagram point Q(((A+B+C)/3),0) point p{((A+B+C)/3),cos(((A+B+C)/3))} from diagram QP>QG cos(((A+B+C)/3))>((cosA+cosB+cosC)/3) 3cos((π/3))>cosA+cosB+cosC cosA+cosB+cosC<(3/2) now maximum value of cosA+cosB+cosC when A=B=C...so A=B=C=(π/3) so cos(π/3)+cos(π/3)+cos(π/3) =(1/2)+(1/2)+(1/2) =(3/2) that means equilateral triangle... [in other cases cosA+cosB+cosC<(3/2)]$
$p o i n t A, B, C l i e s o n c o s x c u r v e$ $p o i n t A c o o r d i n a t e i s (A, c o s A)$ $p o i n t B c o o r d i n a t e (B, c o s B)$ $p o i n t C c o o r d i n a t e (C, c o s C)$ $△ A B C c e n t r o i d i s G$ $c o o r d i n a t e o f G i s (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ $s o G i s (\frac{A + B + C}{3}, \frac{c o s A + c o s B + c o s C}{3})$ $a s p e r d i a g r a m p o i n t Q (\frac{A + B + C}{3}, 0)$ $p o i n t p {\frac{A + B + C}{3}, c o s (\frac{A + B + C}{3})}$ $f r o m d i a g r a m$ $Q P > Q G$ $c o s (\frac{A + B + C}{3}) > \frac{c o s A + c o s B + c o s C}{3}$ $3 c o s (\frac{π}{3}) > c o s A + c o s B + c o s C$ $c o s A + c o s B + c o s C < \frac{3}{2}$ $n o w m a x i m u m v a l u e o f c o s A + c o s B + c o s C$ $w h e n A = B = C . . . s o A = B = C = \frac{π}{3}$ $s o c o s \frac{π}{3} + c o s \frac{π}{3} + c o s \frac{π}{3}$ $= \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$ $= \frac{3}{2} t h a t m e a n s e q u i l a t e r a l t r i a n g l e . . .$ $[i n o t h e r c a s e s c o s A + c o s B + c o s C < \frac{3}{2}]$
Commented by math1967 last updated on 05/Feb/19

$T h a n k y o u s i r$
Commented by math1967 last updated on 05/Feb/19

$o t h e r m e t h o d$ $\cos (A + B) + \cos (B + C) + \cos (C + A) = - \frac{3}{2}$ $\Rightarrow {(\sin A - \sin B)}^{2} + {(\sin B - \sin C)}^{2} + {(\sin C - \sin A)}^{2}$ ${(\cos A + \cos B)}^{2} + {(\cos B + \cos C)}^{2} + {(\cos C + \cos A)}^{2} = 0$ $∴ {(\sin A - \sin B)}^{2} = 0 \Rightarrow A = B$ $s i m i l a r l y B = C ∴ △ A B C e q u i l a t e r a l$ $i f \cos A + \cos B = 0 t h e n A = π - B$ $n o t p o s s i b l e f o r t r i a n g l e$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

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