Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 14130 by Joel577 last updated on 28/May/17

How to find maximum value of k if ((5 − cos 2θ)/(sin θ)) ≥ 2k 0 ≤ θ ≤ π

$$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{if} \\ $$$$\frac{\mathrm{5}\:−\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{sin}\:\theta}\:\:\geqslant\:\mathrm{2}{k}\:\:\:\:\:\:\:\:\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\pi \\ $$

Commented by Joel577 last updated on 28/May/17

((5 − cos 2θ)/(sin θ)) ≥ 2k ((4 + 2sin^2 θ)/(sin θ)) ≥ 2k sin θ is minimum when θ = (π/2) so k = 3 Am I right?

$$\frac{\mathrm{5}\:−\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{sin}\:\theta}\:\geqslant\:\mathrm{2}{k} \\ $$$$\frac{\mathrm{4}\:+\:\mathrm{2sin}^{\mathrm{2}} \:\theta}{\mathrm{sin}\:\theta}\:\geqslant\:\mathrm{2}{k} \\ $$$$\mathrm{sin}\:\theta\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{when}\:\theta\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{so}\:{k}\:=\:\mathrm{3} \\ $$$$\mathrm{Am}\:\mathrm{I}\:\mathrm{right}? \\ $$

Commented by prakash jain last updated on 28/May/17

sin θ is actually maximum when θ=(π/2)

$$\mathrm{sin}\:\theta\:\mathrm{is}\:\mathrm{actually}\:\mathrm{maximum}\:\mathrm{when} \\ $$$$\theta=\frac{\pi}{\mathrm{2}} \\ $$

Commented by prakash jain last updated on 28/May/17

f(θ)=((4 + 2sin^2 θ)/(sin θ)) f′(θ)=cos θ(2−4cosec^2 θ) f′(θ)=0 at θ=(π/2) f′′(θ)=−2sin θ+8cosec^3 θ−4cosec θ f′′((π/2))=−2+8−4>0 f((π/2)) is local minima for f(θ) f((π/2))=((4+2)/1)=6⇒k=3

$${f}\left(\theta\right)=\frac{\mathrm{4}\:+\:\mathrm{2sin}^{\mathrm{2}} \:\theta}{\mathrm{sin}\:\theta}\: \\ $$$${f}'\left(\theta\right)=\mathrm{cos}\:\theta\left(\mathrm{2}−\mathrm{4cosec}^{\mathrm{2}} \theta\right) \\ $$$${f}'\left(\theta\right)=\mathrm{0} \\ $$$${at}\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$${f}''\left(\theta\right)=−\mathrm{2sin}\:\theta+\mathrm{8cosec}^{\mathrm{3}} \theta−\mathrm{4cosec}\:\theta \\ $$$${f}''\left(\frac{\pi}{\mathrm{2}}\right)=−\mathrm{2}+\mathrm{8}−\mathrm{4}>\mathrm{0} \\ $$$${f}\left(\frac{\pi}{\mathrm{2}}\right)\:\mathrm{is}\:\mathrm{local}\:\mathrm{minima}\:\mathrm{for}\:{f}\left(\theta\right) \\ $$$${f}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{4}+\mathrm{2}}{\mathrm{1}}=\mathrm{6}\Rightarrow{k}=\mathrm{3} \\ $$

Commented by Joel577 last updated on 29/May/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Answered by ajfour last updated on 28/May/17

4+(1−cos 2θ)≥2ksin θ 4+2sin^2 θ−2ksin θ≥0 sin^2 θ−ksin θ+2≥0 ⇒for θ=0,π ; k=any real number ⇒ k≤((2+sin^2 θ)/(sin θ)) let f(θ)=((2+sin^2 θ)/(sin θ)) f ′(θ)=((2sin^2 θcos θ−cos θ(2+sin^2 θ))/(sin^2 θ)) =((cos θ(sin^2 θ−2))/(sin^2 θ)) f ′(θ)>0 for (π/2)<θ<π while f ′(θ)<0 for 0<θ<π local minima at x=(π/2) ⇒ f(θ) is minimum for f ′(θ)=0 or cos θ=0, θ=(π/2). f((π/2))=((2+1)/1)=3 k_(max) = minimum of f((π/2)) k_(max) =3 .

$$\mathrm{4}+\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)\geqslant\mathrm{2}{k}\mathrm{sin}\:\theta \\ $$$$\mathrm{4}+\mathrm{2sin}\:^{\mathrm{2}} \theta−\mathrm{2}{k}\mathrm{sin}\:\theta\geqslant\mathrm{0} \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \theta−{k}\mathrm{sin}\:\theta+\mathrm{2}\geqslant\mathrm{0} \\ $$$$\Rightarrow{for}\:\theta=\mathrm{0},\pi\:\:\:;\:\:{k}={any}\:{real}\:{number} \\ $$$$\Rightarrow\:{k}\leqslant\frac{\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:\theta} \\ $$$${let}\:{f}\left(\theta\right)=\frac{\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:{f}\:'\left(\theta\right)=\frac{\mathrm{2sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:\theta−\mathrm{cos}\:\theta\left(\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \theta\right)}{\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\:\:\:\:=\frac{\mathrm{cos}\:\theta\left(\mathrm{sin}\:^{\mathrm{2}} \theta−\mathrm{2}\right)}{\mathrm{sin}^{\mathrm{2}} \:\theta}\: \\ $$$${f}\:'\left(\theta\right)>\mathrm{0}\:{for}\:\frac{\pi}{\mathrm{2}}<\theta<\pi \\ $$$${while}\:{f}\:'\left(\theta\right)<\mathrm{0}\:{for}\:\mathrm{0}<\theta<\pi \\ $$$${local}\:{minima}\:{at}\:{x}=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:{f}\left(\theta\right)\:{is}\:{minimum}\:{for}\:{f}\:'\left(\theta\right)=\mathrm{0} \\ $$$${or}\:\:\mathrm{cos}\:\theta=\mathrm{0},\:\:\theta=\frac{\pi}{\mathrm{2}}. \\ $$$${f}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{2}+\mathrm{1}}{\mathrm{1}}=\mathrm{3} \\ $$$${k}_{{max}} =\:{minimum}\:{of}\:{f}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$${k}_{{max}} =\mathrm{3}\:. \\ $$

Commented by prakash jain last updated on 28/May/17

Please see my comment I think minimum is 6.

$$\mathrm{Please}\:\mathrm{see}\:\mathrm{my}\:\mathrm{comment}\:\mathrm{I}\:\mathrm{think} \\ $$$$\mathrm{minimum}\:\mathrm{is}\:\mathrm{6}. \\ $$

Commented by ajfour last updated on 28/May/17

thank you Mr. Prakash.

$${thank}\:{you}\:{Mr}.\:{Prakash}. \\ $$

Commented by Joel577 last updated on 29/May/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com