nice-calculus-in-AB-C-cos-A-cos-B-cos-C-1-9-find-the-value-of-cos-2-A-cos-2-B-cos-2-C-

Question Number 136094 by mnjuly1970 last updated on 19/Mar/21

\dots . n i c e c a l c u l u s \dots .

i n A \overset{Δ}{B C} :: c o s (A) c o s (B) c o s (C) = \frac{1}{9}

f i n d t h e v a l u e o f :

{c o s}^{2} (A) + {c o s}^{2} (B) + {c o s}^{2} (C) = ?

Commented by MJS_new last updated on 19/Mar/21

\cos α \cos β \cos γ ⩽ \frac{1}{8} \Rightarrow no such triangle exists

Commented by mr W last updated on 19/Mar/21

j u s t b e c a u s e \cos A \cos B \cos C = \frac{1}{9} < \frac{1}{8},

i t s h o u l d b e p o s s i b l e, i t h i n k .

\cos^{2} A + \cos^{2} B + \cos^{2} C = 1,

w i t h A, B, C = (\cos^{- 1} \frac{1}{\sqrt{3}}, \cos^{- 1} \frac{1}{\sqrt{3}}, \cos^{- 1} \frac{1}{3}) \equiv (54.735 °, 54.735 °, 70.529 °)

o r A, B, C = (\cos^{- 1} \frac{1}{\sqrt{6}}, \cos^{- 1} \frac{1}{\sqrt{6}}, \cos^{- 1} \frac{2}{3}) \equiv (65.905 °, 65.905 °, 48.190 °)

Commented by MJS_new last updated on 19/Mar/21

the constant had been \frac{3}{8}, now {it}^{'} s a different

question

Commented by mr W last updated on 19/Mar/21

o k . i {d i d n}^{'} t k n o w t h a t .

Commented by MJS_new last updated on 19/Mar/21

2 {(\frac{1}{\sqrt{3}})}^{2} + {(\frac{1}{3})}^{2} = \frac{7}{9} \neq 1

2 {(\frac{1}{\sqrt{6}})}^{2} + {(\frac{2}{3})}^{2} = \frac{7}{9} \neq 1

Answered by MJS_new last updated on 19/Mar/21

\cos α \cos β \cos (π - α - β) = \frac{1}{9}

- \cos α \cos β \cos (α + β) = \frac{1}{9}

\sin α \cos α \sin β \cos β - \cos^{2} α \cos^{2} β = \frac{1}{9}

- \frac{1 - \tan α \tan β}{(1 + \tan^{2} α) (1 + \tan^{2} β)} = \frac{1}{9}

\cos^{2} α + \cos^{2} β + \cos^{2} (π - α - β) =

= \cos^{2} α + \cos^{2} β + \cos^{2} (α + β) = \dots

\dots = 1 + \frac{2 (1 - \tan α \tan β)}{(1 + \tan^{2} α) (1 + \tan^{2} β)} =

= 1 + 2 \times (- \frac{1}{9}) = \frac{7}{9}

Commented by mr W last updated on 19/Mar/21

b u t i g o t Σ \cos^{2} A = 1 w i t h t h e g i v e n

e x a m p l e s a b o v e . d o e s i t m e a n

t h a t Σ \cos^{2} A \neq c o n s t a n t ?

Commented by MJS_new last updated on 19/Mar/21

{it}^{'} s constant but {it}^{'} s \frac{7}{9} instead of 1

Commented by mr W last updated on 19/Mar/21

i s e e . i m a d e a s i m p l e m i s t a k e .

t h a n k s a l o t!

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