Question

The maximum value of $f\left(x\right)=\sin x(1+\cos x)$ is

• A. $\raisebox{1ex}{$3\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$
• B. $\raisebox{1ex}{$3\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• C. $3\sqrt{3}$
• D. $\sqrt{3}$

Answer 1

The correct option is A

$\raisebox{1ex}{$3\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Step $1:$Solve for the maximum value of $f\left(x\right)=\sin x(1+\cos x)$

$$\begin{gathered} f\left(x\right)=\sin x(1+\cos x) \\ f\text{'}\left(x\right)=\sin x(-\sin x)+(1+\cos x)\cos x \\ =-{\sin }^{2}x+\cos x+{\cos }^{2}x \\ =\cos 2x+\hspace{0.17em}\cos x \\ =2{\cos }^{2}x-1+\cos x \\ =2{\cos }^{2}x+2\cos x-\cos x-1 \\ =2\cos x(\cos x+1)-1(\cos x+1) \\ =(2\cos x-1)(\cos x+1) \end{gathered}$$

Let $f\text{'}\left(x\right)=0$

Therefore $\cos x=\frac{1}{2},\cos x=-1$

The maximum value occurs at $x=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Hence $f\left(\frac{\mathrm{\pi }}{3}\right)=\sin \frac{\mathrm{\pi }}{3}(1+\cos \frac{\mathrm{\pi }}{3})$

$$\begin{gathered} =\left(\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)(1+\frac{1}{2}) \\ =\left(\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2)$}\right. \\ =\frac{3\sqrt{3}}{4} \end{gathered}$$

Hence the maximum value of $f\left(x\right)=\sin x(1+\cos x)$ is $\frac{3\sqrt{3}}{4}$.