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d-dx-1-2-x-2-sin-x-




Question Number 117759 by Dwaipayan Shikari last updated on 13/Oct/20
(dθ/dx)=((√(1−(θ^2 /x^2 )))/(sin(θ+x)))
$$\frac{{d}\theta}{{dx}}=\frac{\sqrt{\mathrm{1}−\frac{\theta^{\mathrm{2}} }{{x}^{\mathrm{2}} }}}{{sin}\left(\theta+{x}\right)} \\ $$
Commented by TANMAY PANACEA last updated on 13/Oct/20
assuming θ and x in radian  so θ=αx  (dθ/dx)=α=((√(1−α^2 ))/(sin(1+α)x))  sin(1+α)x=((√(1−α^2 ))/α)  1≥((√(1−α^2 ))/α)≥−1  1≥((1−α^2 )/α^2 )≥0  2α^2 ≥1→α^2 ≥(1/2)  1−α^2 ≥0→α^2 ≤1  1≥α^2 ≥(1/2)→1≥α≥(1/( (√2)))  x≥αx≥(x/( (√2)))     x≥θ≥(x/( (√2)))  i have just put an idea...
$${assuming}\:\theta\:{and}\:{x}\:{in}\:{radian} \\ $$$${so}\:\theta=\alpha{x} \\ $$$$\frac{{d}\theta}{{dx}}=\alpha=\frac{\sqrt{\mathrm{1}−\alpha^{\mathrm{2}} }}{{sin}\left(\mathrm{1}+\alpha\right){x}} \\ $$$${sin}\left(\mathrm{1}+\alpha\right){x}=\frac{\sqrt{\mathrm{1}−\alpha^{\mathrm{2}} }}{\alpha} \\ $$$$\mathrm{1}\geqslant\frac{\sqrt{\mathrm{1}−\alpha^{\mathrm{2}} }}{\alpha}\geqslant−\mathrm{1} \\ $$$$\mathrm{1}\geqslant\frac{\mathrm{1}−\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$$\mathrm{2}\alpha^{\mathrm{2}} \geqslant\mathrm{1}\rightarrow\alpha^{\mathrm{2}} \geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}−\alpha^{\mathrm{2}} \geqslant\mathrm{0}\rightarrow\alpha^{\mathrm{2}} \leqslant\mathrm{1} \\ $$$$\mathrm{1}\geqslant\alpha^{\mathrm{2}} \geqslant\frac{\mathrm{1}}{\mathrm{2}}\rightarrow\mathrm{1}\geqslant\alpha\geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${x}\geqslant\alpha{x}\geqslant\frac{{x}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:{x}\geqslant\theta\geqslant\frac{{x}}{\:\sqrt{\mathrm{2}}} \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{have}}\:\boldsymbol{{just}}\:\boldsymbol{{put}}\:\boldsymbol{{an}}\:\boldsymbol{{idea}}… \\ $$$$ \\ $$

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