a-3-cos-A-B-b-3-cos-B-C-c-3-cos-A-C-

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