Question
If 0 <\theta < \pi , find the maximum value of \sin \dfrac{\theta }{2} (1+\cos \theta )
If 0 <\theta < \pi , find the maximum value of \sin \dfrac{\theta }{2} (1+\cos \theta )
\because 0 <\theta < \pi
\therefore 0 <\dfrac{ \theta}{2} <\dfrac{ \pi}{2}
\sin \dfrac{ \theta}{2}>0, \cos\dfrac{ \theta}{2} >0
Using the inequality of arithmetic and geometric means
GM \leq AM
\sin \dfrac{\theta }{2} (1+\cos \theta )
=2\sin \dfrac{\theta }{2} \cos^2 \dfrac{\theta }{2}
=\sqrt{2\cdotp2\sin^2 \dfrac{\theta }{2} \cos^2 \dfrac{\theta }{2} \cos^2 \dfrac{\theta }{2} }
\leq \sqrt{2\cdotp \Bigg( \dfrac{ 2\sin^2 \dfrac{\theta }{2}+ \cos^2 \dfrac{\theta }{2} +\cos^2 \dfrac{\theta }{2} }{3} \Bigg) ^3 }
=\sqrt{\dfrac{16}{27} } =\dfrac{4\sqrt{3} }{9}