find the maximum of (1 + cosx) sinx without derivative.


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Question Number 204105 by CrispyXYZ last updated on 06/Feb/24

$find the maximum of (1 + \cos x) \sin x$ $without derivative .$
	Answered by deleteduser1 last updated on 06/Feb/24

	$(1 + c o s x) \frac{\sqrt{3} (s i n x)}{\sqrt{3}} ⩽ {(\frac{1 + c o s (x) + \sqrt{3} s i n x}{2})}^{2} \times \frac{1}{\sqrt{3}}$ $⩽ {(\frac{1 + \sqrt{1^{2} + {(\sqrt{3})}^{2}}}{2})}^{2} \times \frac{1}{\sqrt{3}} = \frac{3 \sqrt{3}}{4}$ $E q u a l i t y h o l d s w h e n c o s x = \sqrt{3} s i n x - 1$ $\Rightarrow {(\sqrt{3} s i n x - 1)}^{2} + {s i n}^{2} x = 1 \Rightarrow 4 {s i n}^{2} x = 2 \sqrt{3} s i n x$ $\Rightarrow s i n (x) = \frac{\sqrt{3}}{2} \Rightarrow x = \frac{π}{3} + 2 n π$
	Commented by deleteduser1 last updated on 06/Feb/24

	$[H o w c = \sqrt{3} w a s g o t t e n a b o v e]$ $\frac{(1 + c o s x) c s i n x}{c} ⩽ {(\frac{1 + c o s x + c s i n x}{2})}^{2} \times \frac{1}{c} ⩽ \frac{1}{c} \frac{{(1 + \sqrt{c^{2} + 1})}^{2}}{4}$ $E q u a l i t y w h e n c o s x = c s i n x - 1$ $\Rightarrow (c^{2} + 1) {s i n}^{2} x - 2 c s i n x = 0 \Rightarrow s i n x = \frac{2 c}{c^{2} + 1}$ $\Rightarrow c o s x = \frac{c^{2} - 1}{c^{2} + 1} \Rightarrow \frac{4 c^{4}}{{(c^{2} + 1)}^{2}} = \frac{2 + c^{2} + 2 \sqrt{c^{2} + 1}}{4}$ $\Rightarrow c = \sqrt{3}$
	Commented by CrispyXYZ last updated on 07/Feb/24

	$N i c e s i r$
	Answered by mr W last updated on 06/Feb/24

	$l e t t = \tan \frac{x}{2}$ $f o r m a x i m u m o f y w e o n l y n e e d t o$ $t r e a t \sin x > 0 o r t > 0 .$ $y = (1 + \cos x) \sin x$ $= (1 + \frac{1 - t^{2}}{1 + t^{2}}) \times \frac{2 t}{1 + t^{2}}$ $= \frac{4 t}{{(1 + t^{2})}^{2}}$ $= \frac{4}{{(\frac{1}{\sqrt{t}} + t \sqrt{t})}^{2}}$ $= \frac{4}{{(\frac{1}{3 \sqrt{t}} + \frac{1}{3 \sqrt{t}} + \frac{1}{3 \sqrt{t}} + t \sqrt{t})}^{2}}$ $⩽ \frac{4}{{(4 \times \sqrt[4]{\frac{1}{3 \sqrt{t}} \times \frac{1}{3 \sqrt{t}} \times \frac{1}{3 \sqrt{t}} \times t \sqrt{t}})}^{2}} = \frac{3 \sqrt{3}}{4}$ $i . e . m a x i m u m i s \frac{3 \sqrt{3}}{4} w h i c h o c c u r s$ $w h e n \frac{1}{3 \sqrt{t}} = t \sqrt{t}, i . e . t = \frac{1}{\sqrt{3}} = \tan \frac{x}{2}$ $o r \frac{x}{2} = k π + \frac{π}{6} o r x = 2 k π + \frac{π}{3}$
	Commented by CrispyXYZ last updated on 07/Feb/24

	$T h a n k s a l o t s i r$
	Answered by CrispyXYZ last updated on 07/Feb/24

	$S o l u t i o n w i t h {J e n s e n}^{'} s I n e q u a l i t y :$ $S = (1 + \cos x) \sin x$ $= \sin x + \frac{1}{2} \sin 2 x$ $T = 2 π$ $I . x \in [0, \frac{π}{2}]$ $S = \sin x + \frac{1}{2} \sin (π - 2 x)$ $= \frac{3}{2} (\frac{2}{3} \sin x + \frac{1}{3} \sin (π - 2 x))$ $⩽ \frac{3}{2} \sin (\frac{2}{3} x + \frac{π}{3} - \frac{2}{3} x) = \frac{3 \sqrt{3}}{4} (x = \frac{π}{3})$ $II . x \in [- π, 0]$ $1 + \cos x ⩾ 0, \sin x ⩽ 0$ $\Rightarrow S ⩽ 0$ $III . x \in [\frac{π}{2}, π]$ $S = \sin (π - x) - \frac{1}{2} \sin (2 π - 2 x)$ $⩽ \sin (π - x) + \frac{1}{2} \sin (2 π - 2 x)$ $< \frac{3 \sqrt{3}}{4} (according to I)$ $Conclusion : S_{m a x} = \frac{3 \sqrt{3}}{4} \Leftrightarrow x = \frac{π}{3} + 2 k π$
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