nice-calculus-In-AB-C-prove-sin-A-2-sin-B-2-sin-C-2-1-8-max-cos-A-2-cos-B-2-cos-C-2-

Question Number 122636 by mnjuly1970 last updated on 18/Nov/20

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

I n A \overset{Δ}{B C} p r o v e ::

* : s i n (\frac{A}{2}) s i n (\frac{B}{2}) s i n (\frac{C}{2}) ⩽ \frac{1}{8}

\dots \dots \dots \dots \dots \dots \dots \dots .

* * :: m a x (c o s (\frac{A}{2}) c o s (\frac{B}{2}) c o s (\frac{C}{2})) = ?

Answered by TANMAY PANACEA last updated on 18/Nov/20

t h r e e p o i n t s A, B a n d C l i e o n c u r v e y = s i n \frac{x}{2}

(A, s i n \frac{A}{2}) (B, s i n \frac{B}{2}) a n d (C, s i n \frac{C}{2})

c e n t r o i d P (\frac{A + B + C}{3}, \frac{s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2}}{3})

o r d i n a t e o f c e n t r o i d P = (\frac{s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2}}{3})

p o i n t Q = {(\frac{A + B + C}{3}) s i n (\frac{\frac{A + B + C}{3}}{2})} l i e s

o n y = s i n (\frac{x}{2})

s i n (\frac{A + B + C}{6}) = o r d i n a t e o f p o i n t Q

s i n (\frac{A + B + C}{6}) > \frac{s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2}}{3}

u s i n g A M > G M

s i n (\frac{180^{o}}{6}) > \frac{s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2}}{3} > {(s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2})}^{\frac{1}{3}}

{(\frac{1}{2})}^{3} > (s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2})

p r o v e d

Commented by TANMAY PANACEA last updated on 18/Nov/20

s a m e m e t h o d f o r c o s θ c u r v e

f i n a l s t e p

c o s (\frac{180^{o}}{6}) > {(c o s \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2})}^{\frac{1}{3}}

{(\frac{\sqrt{3}}{2})}^{3} > (c o s \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2})

\frac{3 \sqrt{3}}{2} > (c o s \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2})

Commented by mnjuly1970 last updated on 18/Nov/20

t h a n k y o u s i r t a n m a y \dots

e x c e l l e n t \dots

Commented by TANMAY PANACEA last updated on 18/Nov/20

m o s t w e l c o m e