Question Number 58025 by Kunal12588 last updated on 16/Apr/19
$${Trace}\:{the}\:{changes}\:{in}\:{the}\:{sign}\:{and}\:{magnitude} \\ $$$${of}\:\:\frac{\mathrm{sin}\:\mathrm{3}\theta}{\mathrm{cos}\:\mathrm{2}\theta}\:{as}\:{the}\:{angle}\:{increases}\:{from}\:\mathrm{0}\:{to}\:\frac{\pi}{\mathrm{2}}. \\ $$$${also}\:{find}\:{its}\:{minimum}\:{and}\:{maximum}\:{values}. \\ $$
Commented by Kunal12588 last updated on 16/Apr/19
Answered by tanmay last updated on 16/Apr/19
![(π/2)≥θ≥0 π≥2θ≥0 ((3π)/2)≥3θ≥0 A) sin3θ=+ve (π/6)≥θ≥0 cos2θ=+ve so ((sin3θ)/(cos2θ))=+ve when (π/6)≥θ≥0 B) sin3θ=+ve [when (π/4)> θ≥(π/6) cos2θ=+ve ((sin3θ)/(cos2θ))=++ve c)sin3θ= +ve when (π/3)≥θ>(π/4) cos2θ=−ve ((sin3θ)/(cos2θ))=−ve d)sin3θ= −ve (π/2)≥θ≥(π/3) cos2θ=−ve so ((sin3θ)/(cos2θ))=+ve pls check... f(θ)=((sin3θ)/(cos2θ)) (df/dθ)=((cos2θ×3cos3θ+2sin3θ×sin2θ)/(cos^2 2θ)) (df/dθ)=((cos2θ×cos3θ+2cos(3θ−2θ))/(cos^2 2θ)) (df/dθ)=((cos3θ×cos2θ+2cosθ)/(cos^2 2θ)) for max/min (df/dθ)=0 cos3θ×cos2θ+2cosθ (4x^3 −3x)(2x^2 −1)+2x 8x^5 −4x^3 −6x^3 +3x+2x 8x^5 −10x^3 +5x x(8x^4 −10x^2 +5) x{2(4x^4 −5x^2 )+5} x[2{(2x^2 )^2 −2×2x^2 ×(5/4)+((25)/(16))−((25)/(16))}+5] x[2(2x^2 −(5/4))^2 −((25)/8)+5] x[2(2x^2 −(5/4))^2 +((15)/8)] so [2(2x^2 −(5/4))^2 +((15)/8)]≠0 hence 8x^5 −10x^3 +5x=0 when x=0 cosθ=0=cos(π/2) [θ=(π/2)] f((π/2))=((sin(((3π)/2)))/(cos(((2π)/2))))=((−1)/(−1))=1(maximum value) f(0)=((sin(3×0))/(cos(2×0)))=0](Q58028.png)
$$\frac{\pi}{\mathrm{2}}\geqslant\theta\geqslant\mathrm{0} \\ $$$$\pi\geqslant\mathrm{2}\theta\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{3}\pi}{\mathrm{2}}\geqslant\mathrm{3}\theta\geqslant\mathrm{0} \\ $$$$\:\: \\ $$$$\left.\:{A}\right)\:\:\:\:{sin}\mathrm{3}\theta=+{ve}\:\:\:\frac{\pi}{\mathrm{6}}\geqslant\theta\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{cos}\mathrm{2}\theta=+{ve}\: \\ $$$$\:\:\:\:\:\:\:\:{so}\:\:\:\frac{{sin}\mathrm{3}\theta}{{cos}\mathrm{2}\theta}=+{ve}\:\:{when}\:\frac{\pi}{\mathrm{6}}\geqslant\theta\geqslant\mathrm{0} \\ $$$$\left.{B}\right)\:\:\:{sin}\mathrm{3}\theta=+{ve}\:\:\:\:\:\:\left[{when}\:\frac{\pi}{\mathrm{4}}>\:\theta\geqslant\frac{\pi}{\mathrm{6}}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{cos}\mathrm{2}\theta=+{ve} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{sin}\mathrm{3}\theta}{{cos}\mathrm{2}\theta}=++{ve} \\ $$$$\left.{c}\right){sin}\mathrm{3}\theta=\:\:\:\:\:+{ve}\:\:\:\:{when}\:\:\frac{\pi}{\mathrm{3}}\geqslant\theta>\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:{cos}\mathrm{2}\theta=−{ve} \\ $$$$\:\:\:\:\:\frac{{sin}\mathrm{3}\theta}{{cos}\mathrm{2}\theta}=−{ve} \\ $$$$\left.{d}\right){sin}\mathrm{3}\theta=\:\:−{ve}\:\:\:\:\:\frac{\pi}{\mathrm{2}}\geqslant\theta\geqslant\frac{\pi}{\mathrm{3}} \\ $$$$\:\:\:\:\:{cos}\mathrm{2}\theta=−{ve} \\ $$$${so}\:\frac{{sin}\mathrm{3}\theta}{{cos}\mathrm{2}\theta}=+{ve} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$$${f}\left(\theta\right)=\frac{{sin}\mathrm{3}\theta}{{cos}\mathrm{2}\theta} \\ $$$$\frac{{df}}{{d}\theta}=\frac{{cos}\mathrm{2}\theta×\mathrm{3}{cos}\mathrm{3}\theta+\mathrm{2}{sin}\mathrm{3}\theta×{sin}\mathrm{2}\theta}{{cos}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$\frac{{df}}{{d}\theta}=\frac{{cos}\mathrm{2}\theta×{cos}\mathrm{3}\theta+\mathrm{2}{cos}\left(\mathrm{3}\theta−\mathrm{2}\theta\right)}{{cos}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$\frac{{df}}{{d}\theta}=\frac{{cos}\mathrm{3}\theta×{cos}\mathrm{2}\theta+\mathrm{2}{cos}\theta}{{cos}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$${for}\:{max}/{min}\:\frac{{df}}{{d}\theta}=\mathrm{0} \\ $$$${cos}\mathrm{3}\theta×{cos}\mathrm{2}\theta+\mathrm{2}{cos}\theta \\ $$$$\left(\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x}\right)\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{2}{x} \\ $$$$\mathrm{8}{x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{2}{x} \\ $$$$\mathrm{8}{x}^{\mathrm{5}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{5}{x} \\ $$$${x}\left(\mathrm{8}{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{5}\right) \\ $$$${x}\left\{\mathrm{2}\left(\mathrm{4}{x}^{\mathrm{4}} −\mathrm{5}{x}^{\mathrm{2}} \right)+\mathrm{5}\right\} \\ $$$${x}\left[\mathrm{2}\left\{\left(\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}×\mathrm{2}{x}^{\mathrm{2}} ×\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{25}}{\mathrm{16}}−\frac{\mathrm{25}}{\mathrm{16}}\right\}+\mathrm{5}\right] \\ $$$${x}\left[\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{8}}+\mathrm{5}\right] \\ $$$${x}\left[\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{15}}{\mathrm{8}}\right] \\ $$$${so}\:\left[\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{15}}{\mathrm{8}}\right]\neq\mathrm{0} \\ $$$${hence}\:\mathrm{8}{x}^{\mathrm{5}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{5}{x}=\mathrm{0}\:{when}\:{x}=\mathrm{0} \\ $$$${cos}\theta=\mathrm{0}={cos}\frac{\pi}{\mathrm{2}}\:\:\left[\theta=\frac{\pi}{\mathrm{2}}\right] \\ $$$${f}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)}{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{2}}\right)}=\frac{−\mathrm{1}}{−\mathrm{1}}=\mathrm{1}\left({maximum}\:{value}\right) \\ $$$${f}\left(\mathrm{0}\right)=\frac{{sin}\left(\mathrm{3}×\mathrm{0}\right)}{{cos}\left(\mathrm{2}×\mathrm{0}\right)}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Kunal12588 last updated on 17/Apr/19
$${thank}\:{you}\:{sir} \\ $$