If-0-lt-x-lt-pi-2-then-prove-sin-cosx-lt-cosx-lt-cos-sinx-not-using-graph-

Question Number 213084 by MATHEMATICSAM last updated on 30/Oct/24

If 0 < x < \frac{π}{2} then prove

\sin (\cos x) < \cos x < \cos (\sin x) .

not using graph .

Answered by issac last updated on 30/Oct/24

x = acos (z), z \in [0, 1]

\sin (\cos (\cos^{- 1} (z))) < \cos (\cos^{- 1} (z)) < \cos (\sin (\cos^{- 1} (z)))

\sin (z) < z < \cos (\sqrt{1 - z^{2}})

\sin (z) < z < \cos (\sqrt{1 - z^{2}}), z \in (0, 1]

\sin (z) \leq z

\frac{\sin (z)}{z} \leq 1 \to \frac{d}{d z} \sin (z) \leq 1 \cos (z) \leq 1, for all z \in [0, 1]

\sin (z) \leq z \leq \cos (\sqrt{1 - z^{2}})

z \leq \cos (\sqrt{1 - z^{2}})

e^{z} = 1 + z + \frac{1}{2!} z^{2} + \frac{1}{3!} z^{3} \dots . . (Tailer series)

Couchy - Schwartz Inequality

{(a_{1} z_{1} + a_{2} z_{2} + \dots \dots . a_{n} z_{n})}^{2} \leq (a_{1}^{2} + a_{2}^{2} + \dots a_{n}^{2}) (z_{1}^{2} + z_{2}^{2} + \dots z_{n}^{2})

{(1 + z + \frac{1}{2!} z^{2} + \frac{1}{3!} z^{3} \dots . .)}^{2} \leq (1^{2} + 1^{2} + {(\frac{1}{2!})}^{2} + {(\frac{1}{3!})}^{2} + \dots) (1^{2} + z^{2} + z^{4} + \dots .)

e^{2 z} \leq \frac{I_{0} (2)}{1 - z^{2}}, I_{0} (2) \approx 2.2796 \dots .

I_{ν} (x) is Modified 1 st Bessel function

e^{z} \leq \frac{I_{0} (2)}{1 - \frac{1}{4} z^{2}} \to e^{z} \leq \frac{4 \cdot I_{0} (2)}{4 - z^{2}}, z \in [0, 1]

e^{i z} + e^{- i z} \leq \frac{2 I_{0} (2)}{\frac{1}{4} z^{2} + 1} \to \cos (z) \leq \frac{4 \cdot I_{0} (2)}{z^{2} + 4}

\cos (\sqrt{1 - z^{2}}) \leq \frac{4 I_{0} (2)}{5 - z^{2}}

z \leq \frac{4 I_{0} (2)}{5 - z^{2}} \to 5 z - z^{3} \leq 4 I_{0} (2)

0 \leq z^{3} - 5 z + 4 I_{0} (2) satisfy for all z \in [0, 1]

Q . E . D

Commented by issac last updated on 30/Oct/24

I use Couchy - schwartz inequality

and Manipulated inequality

if A, B is positive and k is positive

if A \leq B, A + k \leq B + k is true

S o, I use that fact :)

Answered by mr W last updated on 30/Oct/24

f o r 0 < x < \frac{π}{2} :

\sin x < x \Rightarrow \cos (\sin x) > \cos x ✓

0 < \cos x < 1 < \frac{π}{2} \Rightarrow \sin (\cos x) < \cos x ✓

Commented by mr W last updated on 30/Oct/24

I u s e m y b r a i n o n l y :)

Commented by Frix last updated on 31/Oct/24

{It}^{'} s good you use y o u r brain only because

brain rents had been raised lately .

Commented by mr W last updated on 31/Oct/24

i^{'} v e g o t a f l a t r a t e :)

Answered by MrGaster last updated on 30/Oct/24

\sin (\cos x) < \cos x \Leftrightarrow \sin (\cos x) - \cos x < 0 \Leftrightarrow 2 \cos (\frac{\cos x + \frac{π}{2} - x}{2}) \sin (\frac{\cos x - (\frac{π}{2} - x)}{2}) < 0 \Leftrightarrow 2 \cos (\frac{π}{4} + \frac{x - \cos x}{2}) \sin (\frac{\cos x + x - \frac{π}{2}}{2}) < 0

\cos x < \cos (\sin x) \Leftrightarrow \cos x - \cos (\sin x) < 0 \Leftrightarrow - 2 \sin (\frac{x + \sin x}{2}) \sin (\frac{x - \sin x}{2}) < 0 \Leftrightarrow 2 \sin (\frac{x + \sin x}{2}) \sin (\frac{\sin x - x}{2}) > 0

[Q . E . D]

Answered by Berbere last updated on 31/Oct/24

s i n (x) ⩽ x

c o s (x) ⩽ 1 \Rightarrow \int_{0}^{x} c o s (t) ⩽ \int_{0}^{x} 1 d t \Rightarrow s i n (x) ⩽ x \Rightarrow s i n (c o s (x)) ⩽ c o s (x)

c o s (s i n (x)) > c o s (x)

s i n (x) < x b y p r o v i o u s

c o s d e c r e a s e \Rightarrow c o s (s i n (x)) ⩾ c o s (x)

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