Question Number 7055 by Tawakalitu. last updated on 08/Aug/16
![Show that ∀ x ∈ [ 0, (Π/2) ] , tanx + sin4x ≥ 0](Q7055.png)
$${Show}\:{that}\:\forall\:{x}\:\in\:\left[\:\mathrm{0},\:\frac{\Pi}{\mathrm{2}}\:\right]\:,\:{tanx}\:+\:{sin}\mathrm{4}{x}\:\geqslant\:\mathrm{0}\: \\ $$$$ \\ $$
Commented by Yozzii last updated on 08/Aug/16
![If 0≤x≤(π/2)⇒0≤4x≤2π For −1≤sin4x<0⇒π<4x≤((3π)/2) or ((3π)/2)≤4x<2π ⇒(π/4)<x≤((3π)/8) or ((3π)/8)≤x<(π/2) ⇒1<tanx≤tan((3π)/8) or tan((3π)/8)≤tanx<+∞ ⇒tanx>1 ∀x∈((π/4),(π/2)) ∴tanx+sin4x>1+sin4x≥1+(−1)=0 ⇒tanx+sin4x>0 if x∈((π/4),(π/2)). If 0≤sin4x≤1⇒ 0≤4x≤(π/2) or (π/2)≤4x≤π ⇒0≤x≤(π/8) or (π/8)≤x≤(π/4) ⇒0≤tanx≤tan(π/8) or tan(π/8)≤tanx≤1 ⇒tanx≥0⇒tanx+sin4x≥sin4x≥0 ⇒tanx+sin4x≥0 ∀x∈[0,π/4] ∴ ∀x∈[0,π/2], tanx+sin4x≥0.](Q7057.png)
$${If}\:\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}\Rightarrow\mathrm{0}\leqslant\mathrm{4}{x}\leqslant\mathrm{2}\pi \\ $$$${For}\:−\mathrm{1}\leqslant{sin}\mathrm{4}{x}<\mathrm{0}\Rightarrow\pi<\mathrm{4}{x}\leqslant\frac{\mathrm{3}\pi}{\mathrm{2}}\:{or}\:\frac{\mathrm{3}\pi}{\mathrm{2}}\leqslant\mathrm{4}{x}<\mathrm{2}\pi\: \\ $$$$\Rightarrow\frac{\pi}{\mathrm{4}}<{x}\leqslant\frac{\mathrm{3}\pi}{\mathrm{8}}\:{or}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\leqslant{x}<\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{1}<{tanx}\leqslant{tan}\frac{\mathrm{3}\pi}{\mathrm{8}}\:{or}\:{tan}\frac{\mathrm{3}\pi}{\mathrm{8}}\leqslant{tanx}<+\infty \\ $$$$\Rightarrow{tanx}>\mathrm{1}\:\forall{x}\in\left(\frac{\pi}{\mathrm{4}},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\therefore{tanx}+{sin}\mathrm{4}{x}>\mathrm{1}+{sin}\mathrm{4}{x}\geqslant\mathrm{1}+\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{tanx}+{sin}\mathrm{4}{x}>\mathrm{0}\:{if}\:{x}\in\left(\frac{\pi}{\mathrm{4}},\frac{\pi}{\mathrm{2}}\right). \\ $$$${If}\:\mathrm{0}\leqslant{sin}\mathrm{4}{x}\leqslant\mathrm{1}\Rightarrow\:\mathrm{0}\leqslant\mathrm{4}{x}\leqslant\frac{\pi}{\mathrm{2}}\:{or}\:\frac{\pi}{\mathrm{2}}\leqslant\mathrm{4}{x}\leqslant\pi \\ $$$$\Rightarrow\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{8}}\:{or}\:\frac{\pi}{\mathrm{8}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{0}\leqslant{tanx}\leqslant{tan}\frac{\pi}{\mathrm{8}}\:{or}\:{tan}\frac{\pi}{\mathrm{8}}\leqslant{tanx}\leqslant\mathrm{1} \\ $$$$\Rightarrow{tanx}\geqslant\mathrm{0}\Rightarrow{tanx}+{sin}\mathrm{4}{x}\geqslant{sin}\mathrm{4}{x}\geqslant\mathrm{0} \\ $$$$\Rightarrow{tanx}+{sin}\mathrm{4}{x}\geqslant\mathrm{0}\:\:\forall{x}\in\left[\mathrm{0},\pi/\mathrm{4}\right] \\ $$$$\therefore\:\forall{x}\in\left[\mathrm{0},\pi/\mathrm{2}\right],\:{tanx}+{sin}\mathrm{4}{x}\geqslant\mathrm{0}. \\ $$
Commented by Tawakalitu. last updated on 08/Aug/16
$${Thank}\:{you}\:{so}\:{much}\:{sir}.\:{i}\:{really}\:{appreiate}. \\ $$
Answered by Yozzii last updated on 08/Aug/16
$${Check}\:{comments}\:{for}\:{an}\:{answer}. \\ $$