a^3 cos(A−B)+b^3 cos(B−C)+c^3 cos(A−C)=?


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Question Number 212232 by Davidtim last updated on 07/Oct/24

$a^{3} c o s (A - B) + b^{3} c o s (B - C) + c^{3} c o s (A - C) = ?$
Commented by Davidtim last updated on 07/Oct/24

$I k i n d l y r e q u e s t y o u f o r h e l p i n g$
	Answered by Ghisom last updated on 08/Oct/24

	$this makes no sense . it should be$ $a^{3} \cos (B - C) + b^{3} \cos (C - A) + c^{3} \cos (A - B)$ $which gives$ $3 a b c$ $use$ $\cos (x - y) = \cos x \cos y + \sin x \sin y$ $and$ $a^{2} + b^{2} - 2 a b \cos C = c^{2}$ $\Rightarrow$ $\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$ $\sin C = \frac{\sqrt{(a + b + c) (a + b - c) (a + c - b) (b + c - a)}}{2 a b}$ $[for the other angles rotate a \to b \to c \to a]$
	Commented by Davidtim last updated on 08/Oct/24

	$b y w h i c h l a w y o u h a v e w r i t e n t h e r e l a t i o n$ $f o r s i n C ? j u s t a l i t l e b i t d e s c r i b e f o r m e t o k n o w .$
	Commented by mr W last updated on 08/Oct/24

	$Δ = \frac{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}{4}$ $Δ = \frac{a b \sin C}{2}$ $\frac{a b \sin C}{2} = \frac{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}{4}$ $\Rightarrow \sin C = \frac{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}{2 a b}$
	Commented by Davidtim last updated on 08/Oct/24

	$s o r r y, w h a t d o e s Δ m o d i f y h e r e ?$
	Commented by Davidtim last updated on 08/Oct/24

	$s o r r y d e a r, w h a t d o e s m l d i f y Δ h e r e ?$ $b e c a u s e h e r e i s n o a n y e q u a t i o n ?$ $o r {i t}^{'} s a r e a o f t r i a n g l e ?$
	Commented by mr W last updated on 08/Oct/24

	$y o u r q u e s t i o n i s a b o u t t r i s n g l e, s o i$ $g u e s s y o u k n o w w h a t Δ m e a n s h e r e,$ $e s p e c i a l l y i n$ $Δ = \frac{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}{4}$ $t h i s i s t h e f a m o u s {h e r o n}^{'} s f o r m u l a$ $f o r a r e a o f t r i a n g l e .$
	Commented by Frix last updated on 08/Oct/24
	https://en.m.wikipedia.org/wiki/Heron%27s_formula
	Commented by Davidtim last updated on 08/Oct/24

	$B u t t h e {h e r o n}^{'} s f o r m u l a i s b e l l o w :$ $Δ = \sqrt{p (p - a) (p - b) (p - c)}$
	Commented by Davidtim last updated on 08/Oct/24

	$p = \frac{a + b + c}{2}$
	Commented by Frix last updated on 08/Oct/24

	$Use your brain and simple algebra :$ $Δ^{2} = p (p - a) (p - b) (p - c) =$ $= \frac{a + b + c}{2} (\frac{a + b + c}{2} - a) (\frac{a + b + c}{2} - b) (\frac{a + b + c}{2} - c) =$ $= \frac{(a + b + c) (b + c - a) (a + c - b) (a + b - c)}{16}$ $Δ = \frac{\sqrt{(a + b + c) (a + b - c) (a + c - b) (b + c - a)}}{4}$
	Commented by Davidtim last updated on 08/Oct/24

	$o r s i r v e r y n i c e$
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