Given triangle ABC, what is the maximum value of y=2cosA + cosB + cosC?


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Question Number 137038 by bobhans last updated on 29/Mar/21

Given triangle ABC, what is the maximum value of y=2cosA + cosB + cosC?
	Answered by mr W last updated on 29/Mar/21

	$A = π - (B + C)$ $y = - 2 \cos (B + C) + \cos B + \cos C$ $d u e t o s y m m e t r y : B = C = x$ $y = - 2 \cos 2 x + 2 \cos x$ $y = - 4 \cos^{2} x + 2 + 2 \cos x)$ $y = \frac{9}{4} - (4 \cos^{2} x - 2 \cos x + \frac{1}{4})$ $y = \frac{9}{4} - {(2 \cos x - \frac{1}{2})}^{2}$ $y_{m a x} = \frac{9}{4} w h e n B = C = x = \cos^{- 1} \frac{1}{4}$
	Commented by bobhans last updated on 29/Mar/21

	$why B and C symetri sir$
	Commented by mr W last updated on 29/Mar/21

	$i n t h e f u n c t i o n$ $y = - 2 \cos (B + C) + \cos B + \cos C$ $y o u c a n e x c h a n g e B a n d C a n d t h e$ $f u n c t i o n r e m a i n s t h e s a m e . w h e n$ $s u c h a f u n c t i o n h a s m a x i m u m o r$ $m i n i m u m, t h e n B = C .$
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