Show-that-x-0-2-tanx-sin4x-0-

Question Number 7055 by Tawakalitu. last updated on 08/Aug/16

S h o w t h a t \forall x \in [0, \frac{Π}{2}], t a n x + s i n 4 x ⩾ 0

Commented by Yozzii last updated on 08/Aug/16

I f 0 ⩽ x ⩽ \frac{π}{2} \Rightarrow 0 ⩽ 4 x ⩽ 2 π

F o r - 1 ⩽ s i n 4 x < 0 \Rightarrow π < 4 x ⩽ \frac{3 π}{2} o r \frac{3 π}{2} ⩽ 4 x < 2 π

\Rightarrow \frac{π}{4} < x ⩽ \frac{3 π}{8} o r \frac{3 π}{8} ⩽ x < \frac{π}{2}

\Rightarrow 1 < t a n x ⩽ t a n \frac{3 π}{8} o r t a n \frac{3 π}{8} ⩽ t a n x < + \infty

\Rightarrow t a n x > 1 \forall x \in (\frac{π}{4}, \frac{π}{2})

∴ t a n x + s i n 4 x > 1 + s i n 4 x ⩾ 1 + (- 1) = 0

\Rightarrow t a n x + s i n 4 x > 0 i f x \in (\frac{π}{4}, \frac{π}{2}) .

I f 0 ⩽ s i n 4 x ⩽ 1 \Rightarrow 0 ⩽ 4 x ⩽ \frac{π}{2} o r \frac{π}{2} ⩽ 4 x ⩽ π

\Rightarrow 0 ⩽ x ⩽ \frac{π}{8} o r \frac{π}{8} ⩽ x ⩽ \frac{π}{4}

\Rightarrow 0 ⩽ t a n x ⩽ t a n \frac{π}{8} o r t a n \frac{π}{8} ⩽ t a n x ⩽ 1

\Rightarrow t a n x ⩾ 0 \Rightarrow t a n x + s i n 4 x ⩾ s i n 4 x ⩾ 0

\Rightarrow t a n x + s i n 4 x ⩾ 0 \forall x \in [0, π / 4]

∴ \forall x \in [0, π / 2], t a n x + s i n 4 x ⩾ 0 .

Commented by Tawakalitu. last updated on 08/Aug/16

T h a n k y o u s o m u c h s i r . i r e a l l y a p p r e i a t e .

Answered by Yozzii last updated on 08/Aug/16

C h e c k c o m m e n t s f o r a n a n s w e r .

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