In a ΔABC prove that: ((sin A + sin B)/2) ≤ sin (((A + B)/2))


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Question Number 13201 by Tinkutara last updated on 16/May/17

$In a Δ A B C prove that :$ $\frac{\sin A + \sin B}{2} ⩽ \sin (\frac{A + B}{2})$
Commented by ajfour last updated on 16/May/17

$k i n d l y c h e c k i f$ $B = 0 a n d A = - \frac{π}{3} .$
	Answered by ajfour last updated on 16/May/17

	$A + B = π - C$ $\frac{A + B}{2} = \frac{π}{2} - \frac{C}{2}$ $\Rightarrow \sin (\frac{A + B}{2}) = \cos \frac{C}{2} > 0$ $\sin A + \sin B = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})$ $\frac{\sin A + \sin B}{2 \sin (\frac{A + B}{2})} = \cos (\frac{A - B}{2}) ⩽ 1$ $h e n c e$ $\frac{\sin A + \sin B}{2} ⩽ \sin (\frac{A + B}{2})$ $\sin (\frac{A + B}{2}) > 0 s o i n e q u a l i t y$ ${d o e s n}^{'} t c h a n g e .$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

	$s i n (\frac{A + B}{2}) = s i n (90 - \frac{C}{2}) = c o s \frac{C}{2}$ $s i n A + s i n B = 2 s i n \frac{A + B}{2} c o s \frac{A - B}{2} =$ $= 2 s i n (90 - \frac{C}{2}) c o s (\frac{A - B}{2}) = 2 c o s \frac{C}{2} c o s \frac{A - B}{2}$ $\Rightarrow \frac{2 c o s \frac{C}{2} c o s \frac{A - B}{2}}{2} ⩽ c o s \frac{C}{2} \Rightarrow$ $\Rightarrow c o s \frac{A - B}{2} ⩽ 1 (a l w y e s t r u e) \forall x \in R : c o s x ⩽ 1$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

	$s i n A + s i n B = \frac{a}{2 R} + \frac{b}{2 R} = \frac{a + b}{2 R}$ $s i n (\frac{A + B}{2}) = c o s \frac{C}{2} = \sqrt{\frac{1 + c o s C}{2}} =$ $= \sqrt{\frac{1 + \frac{a^{2} + b^{2} - c^{2}}{2 a b}}{2}} = \sqrt{\frac{(a + b + c) (a + b - c)}{4 a b}}$ $= \sqrt{\frac{p (p - c)}{a b}} (p = \frac{a + b + c}{2}, S = \frac{a b c}{4 R})$ $\Rightarrow \frac{a + b}{4 R} ⩽ \sqrt{\frac{p (p - c)}{a b}} \Rightarrow \frac{{(a + b)}^{2}}{16 R^{2}} ⩽ \frac{p (p - c)}{a b} \Rightarrow$ $\frac{{(a + b)}^{2}}{\frac{a^{2} b^{2} c^{2}}{S^{2}}} ⩽ \frac{p (p - c)}{a b} \Rightarrow S^{2} . \frac{{(a + b)}^{2}}{{a b c}^{2}} ⩽ p (p - c) \Rightarrow$ $p (p - a) (p - b) (p - c) \frac{{(a + b)}^{2}}{{a b c}^{2}} ⩽ p (p - c) \Rightarrow$ $(p - a) (p - b) {(a + b)}^{2} ⩽ {a b c}^{2} \Rightarrow$ $(c + b - a) (c - b + a) {(a + b)}^{2} ⩽ 4 {a b c}^{2} \Rightarrow$ $(c^{2} - {(b - a)}^{2}) {(a + b)}^{2} ⩽ 4 {a b c}^{2} \Rightarrow$ $c^{2} ({(a + b)}^{2} - 4 a b) ⩽ {(b - a)}^{2} {(b + a)}^{2} \Rightarrow$ $c^{2} {(b - a)}^{2} ⩽ {(b - a)}^{2} {(b + a)}^{2} \Rightarrow c^{2} ⩽ {(b + a)}^{2} \Rightarrow$ $c ⩽ b + a . ◼ (p r o v e d)$ $n o t e : i f a = b \Rightarrow \frac{s i n A + s i n B}{2} = s i n A ⩽ s i n (\frac{A + A}{2}) = s i n A .$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

	$w e c a n p r o v e t h a t :$ $\frac{s i n A + s i n B + s i n C}{3} ⩽ s i n (\frac{A + B + C}{3}) = \frac{\sqrt{3}}{2}$ $s i n A + s i n B + s i n C ⩽ \frac{3 \sqrt{3}}{2}$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

	${(\frac{s i n A + s i n B}{2})}^{2} ⩽ {s i n}^{2} (\frac{A + B}{2}) = {s i n}^{2} (90 - \frac{C}{2}) = {c o s}^{2} \frac{C}{2}$ ${s i n}^{2} A + 2 s i n A s i n B + {s i n}^{2} B ⩽ 2 (1 + c o s C) = 2 (1 + c o s (180 - A - B))$ $1 - {c o s}^{2} A + 2 s i n A s i n B + 1 - {c o s}^{2} B ⩽ 2 - 2 c o s A . c o s B + 2 s i n A . s i n B$ $- {c o s}^{2} A + 2 c o s A . c o s B - {c o s}^{2} B ⩽ 0$ ${c o s}^{2} A - 2 c o s A . c o s B + {c o s}^{2} B ⩾ 0$ ${(c o s A - c o s B)}^{2} ⩾ 0 . ◼ (p r o v e d)$
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