find-the-maximum-of-1-cosx-sinx-without-derivative-

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

find the maximum of (1 + \cos x) \sin x

without derivative .

Answered by AST 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 AST 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