Question

If \dfrac{3\pi}{2} < \alpha <2\pi , simplify \sqrt{\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\dfrac{1}{2} +\dfrac{1}{2}\cos 2\alpha } }

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer

\because \dfrac{3\pi}{2} < \alpha <2\pi

\therefore \cos \alpha>0


\because \dfrac{3\pi}{4} <\dfrac{ \alpha}{2} <\pi

\therefore \cos \dfrac{ \alpha}{2} < 0


\sqrt{\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\dfrac{1}{2} +\dfrac{1}{2}\cos 2\alpha } }


= \sqrt{\dfrac{1}{2}+\dfrac{1}{2 }\sqrt{\dfrac{1+\cos 2\alpha}{2} }}


= \sqrt{\dfrac{1}{2}+\dfrac{1}{2 }\sqrt{\dfrac{1+2\cos^2 \alpha -1}{2} }}

= \sqrt{\dfrac{1}{2}+\dfrac{1}{2}\cos \alpha }

=\sqrt{\dfrac{1}{2}(1+\cos \alpha ) }

=\sqrt{\dfrac{1}{2}\cdotp 2\cos^2 \dfrac{\alpha }{2} }

=-\cos \dfrac{\alpha }{2}

Steven Zheng posted 1 month ago

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