Find-the-maximum-of-f-x-cos-x-tan-x-1-tan-x-sin-x-in-terms-of-with-gt-1-and-0-lt-x-lt-tan-1-

Question Number 48739 by mr W last updated on 28/Nov/18

F i n d t h e m a x i m u m o f

f (x) = \cos x + \frac{λ \tan x - 1}{\tan x - λ} \sin x

i n t e r m s o f λ

w i t h λ > 1 a n d 0 < x < \tan^{- 1} λ

Answered by MJS last updated on 28/Nov/18

f (x) = λ \frac{\cos 2 x}{λ \cos x - \sin x}

x = \arctan t \Rightarrow \cos x = \frac{1}{\sqrt{t^{2} + 1}}; \sin x = \frac{t}{\sqrt{t^{2} + 1}}

f (t) = λ \frac{(t - 1) (t + 1)}{(t - λ) \sqrt{t^{2} + 1}}

f^{'} (t) = - λ \frac{λ t^{3} - 3 t^{2} + 3 λ t - 1}{(t - λ) {(t^{2} + 1)}^{3 / 2}}

t^{3} - \frac{3}{λ} t^{2} + 3 t - \frac{1}{λ} = 0

t = z + \frac{1}{λ}

z^{3} + \frac{3 (λ^{2} - 1)}{λ^{2}} z + \frac{2 (λ^{2} - 1)}{λ^{3}} = 0

D = \frac{{(λ^{2} - 1)}^{2}}{λ^{4}} ⩾ 0 ∣ λ \in R

z_{1} = \frac{\sqrt[3]{λ^{2} - 1}}{λ} (\sqrt[3]{λ - 1} - \sqrt[3]{λ + 1})

t_{1} = z_{1} + \frac{1}{λ} = \frac{1}{λ} (1 + \sqrt[3]{λ^{2} - 1} (\sqrt[3]{λ - 1} - \sqrt[3]{λ + 1}))

x_{1} = \arctan (\frac{1}{λ} (1 + \sqrt[3]{λ^{2} - 1} (\sqrt[3]{λ - 1} - \sqrt[3]{λ + 1})))

Commented by mr W last updated on 28/Nov/18

t h a n k y o u s i r!

{i t}^{'} s c o r r e c t a n d p e r f e c t! g r e a t j o b!

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