Question

If 0 <\theta < \pi , find the maximum value of \sin \dfrac{\theta }{2} (1+\cos \theta )

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer

\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}

Steven Zheng posted 1 month ago

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