Question

Find the maximum value of the function

y = \sin (\dfrac{\pi}{3}-2\theta )+\cos (\dfrac{\pi}{3}+2\theta )

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago


Answer

y = \sin (\dfrac{\pi}{3}-2\theta )+\cos (\dfrac{\pi}{3}+2\theta )

=\sin \dfrac{\pi}{3} \cos 2\theta -\cos \dfrac{\pi}{3}\sin 2\theta +\cos \dfrac{\pi}{3}\cos 2\theta -\sin \dfrac{\pi}{3}\sin 2\theta

=\dfrac{\sqrt{3} }{2} \cos 2\theta - \dfrac{1}{2}\sin 2\theta +\dfrac{1}{2}\cos 2\theta - \dfrac{\sqrt{3} }{2} \sin 2\theta

=(\dfrac{\sqrt{3} }{2}+ \dfrac{1}{2}) \cos 2\theta - (\dfrac{\sqrt{3} }{2}+ \dfrac{1}{2}) \sin 2\theta

=(\dfrac{\sqrt{3} +1 }{2}) (\cos 2\theta - \sin 2\theta)

=(\dfrac{\sqrt{3} +1 }{2})\cdotp \sqrt{2} \cdotp \dfrac{\cos 2\theta - \sin 2\theta}{\sqrt{2} }

=\dfrac{\sqrt{6}+\sqrt{2} }{2} (\sin \dfrac{\pi}{4} \cos 2\theta - \cos \dfrac{\pi}{4} \sin 2\theta)

=-\dfrac{\sqrt{6}+\sqrt{2} }{2} \sin ( 2\theta - \dfrac{\pi}{4} )

When 2\theta- \dfrac{\pi}{4} = 2k\pi-\dfrac{\pi}{2} ,

\theta = k\pi - \dfrac{\pi}{8} , in which k\in Z

The maximum value of the function is \dfrac{\sqrt{6}+\sqrt{2} }{2} and the period is \pi

Steven Zheng posted 2 months ago

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