Prove-the-following-inequalities-hold-true-x-R-a-cos-cosx-gt-0-b-cos-sinx-gt-sin-cosx-

Question Number 119246 by 1549442205PVT last updated on 23/Oct/20

Prove the following inequalities hold

true \forall x \in R

a) \cos (cosx) > 0

b) \cos (sinx) > \sin (cosx)

Answered by Olaf last updated on 23/Oct/20

a)

- \frac{π}{2} < - 1 ⩽ \cos x ⩽ + 1 < + \frac{π}{2}

\Rightarrow \cos (\cos x) > 0 (trivial)

Answered by mindispower last updated on 23/Oct/20

c o s (s i n (x)) > 0, \forall x \in R

s i n (c o s (x)) < 0, \forall x \in [- π, - \frac{π}{2}] \cup [\frac{π}{2}, π]

s o w e w o r c k j u s t i n [- \frac{π}{2}, \frac{π}{2}]

x \to c o s (s i n (- x)) = c o s (s i n (- x))

s i n (c o s (- x)) = s i n (c o s (x)) \Rightarrow

j u s t x \in [0, \frac{π}{2}]

l e t s [s o l v e i n x \in [0, \frac{π}{2}]

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

\Leftrightarrow

s i n (\frac{π}{2} - s i n (x)) > s i n (c o s (x))

s i n c e c o s (x), \frac{π}{2} - s i n (x) \in [0, \frac{π}{2}] a n d s i n i n c r e a s e

f u n c t i o n

\Leftrightarrow \frac{π}{2} - s i n (x) > c o s (x)

\Leftrightarrow s i n (x) + c o s (x) < \frac{π}{2} \dots E

∣ s i n (x) + c o s (x) ∣⩽ \sqrt{1^{2} + 1^{2}} . \sqrt{{c o s}^{2} (x) + {s i n}^{2}} = \sqrt{2} < \frac{π}{2}

c a u c h y s h w a r t z . .

b y e q u i v a l e n t E t r u e

\Rightarrow s i n (\frac{π}{2} - s i n (x)) > s i n (c o s (x))

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