Determine-the-maximum-value-of-1-cos-x-sin-x-cos-x-2-where-x-ranges-over-all-real-numbers-

Question Number 116374 by bemath last updated on 03/Oct/20

Determine the maximum value of

\frac{1 + \cos x}{\sin x + \cos x + 2} where x ranges over all

real numbers .

Answered by MJS_new last updated on 03/Oct/20

x = 2 \arctan t

\frac{1 + \cos x}{\sin x + \cos x + 2} = \frac{2}{t^{2} + 2 t + 3}

f (t) = t^{2} + 2 t + 3 = 0 \Rightarrow t = - 1 \pm \sqrt{\dots} \Rightarrow

minimum of f (t) is at t = - 1

\Rightarrow

maximum of \frac{1 + \cos x}{\sin x + \cos x + 2} is 1

Commented by bemath last updated on 03/Oct/20

thank you prof

Answered by john santu last updated on 03/Oct/20

L e t f (x) = \frac{1 + \cos x}{\sin x + 1 + \cos x + 1}

\Rightarrow f (x) = \frac{1}{1 + \frac{1 + \sin x}{1 + \cos x}} .

L e t t i n g u = \frac{1 + \sin x}{1 + \cos x}, i t c l e a r t h a t

u ⩾ 0 a n d s o f (x) ⩽ 1 w h e r e t h e

e q u a l i t y h o l d s w h e n u = 0 . T h u s

t h e m a x i m u m v a l u e o f y i s 1 w h e n

\sin x = - 1 .