Question Number 174271 by mnjuly1970 last updated on 28/Jul/22
![max ( sin(x).cos^( 3) (x))=? x∈ [0 , (π/2) ]](https://www.tinkutara.com/question/Q174271.png)
$$ \\ $$$$\:\:\:\:{max}\:\left(\:\:{sin}\left({x}\right).{cos}^{\:\mathrm{3}} \left({x}\right)\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:{x}\in\:\left[\mathrm{0}\:,\:\frac{\pi}{\mathrm{2}}\:\right] \\ $$
Commented by blackmamba last updated on 28/Jul/22
![f(x)=sin x(1−sin^2 x)cos x f(x)=(1/2)sin 2x(1−(((1−cos 2x)/2))) f(x)=(1/4)sin 2x(1+cos 2x) f(x)=(1/4)sin 2x+(1/8)sin 4x f ′(x)=(1/2)cos 2x+(1/2)cos 4x=0 ⇒2cos 3x cos x = 0 { ((cos 3x=0⇒3x=(π/2); x=(π/6))),((cos x=0⇒x=(π/2))) :} { ((for x=(π/6)⇒f((π/6))=(1/4)sin ((π/3))[1+cos ((π/3))])),((for x=(π/2)⇒f((π/2))=(1/4)sin π[1+cos π] = 0)) :} max = f((π/6))=(1/4)×(1/2)(√3) ×(3/2)=((3(√3))/(16))](https://www.tinkutara.com/question/Q174275.png)
$${f}\left({x}\right)=\mathrm{sin}\:{x}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)\mathrm{cos}\:{x} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\left(\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}\right)\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}{x}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}\:\mathrm{4}{x} \\ $$$${f}\:'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{4}{x}=\mathrm{0} \\ $$$$\:\Rightarrow\mathrm{2cos}\:\mathrm{3}{x}\:\mathrm{cos}\:{x}\:=\:\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{cos}\:\mathrm{3}{x}=\mathrm{0}\Rightarrow\mathrm{3}{x}=\frac{\pi}{\mathrm{2}};\:{x}=\frac{\pi}{\mathrm{6}}}\\{\mathrm{cos}\:{x}=\mathrm{0}\Rightarrow{x}=\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\:\begin{cases}{{for}\:{x}=\frac{\pi}{\mathrm{6}}\Rightarrow{f}\left(\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}\right)\left[\mathrm{1}+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}\right)\right]}\\{{for}\:{x}=\frac{\pi}{\mathrm{2}}\Rightarrow{f}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\pi\left[\mathrm{1}+\mathrm{cos}\:\pi\right]\:=\:\mathrm{0}}\end{cases} \\ $$$${max}\:=\:{f}\left(\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{16}} \\ $$
Commented by mnjuly1970 last updated on 28/Jul/22
$${thx}\:{alot}\:\:{Sir} \\ $$