Question

Does there exist \alpha \in (0,\dfrac{\pi}{2}) such that the functions x^2-4x\cos \alpha +2 = 0 and x^2-4x\sin \alpha -2=0 in terms of x have an equal real root? If there exists, find the value of \alpha . If not, explain why.

Collected in the board: Trigonometry

Steven Zheng posted 5 months ago

Answer

x^2-4x\cos \alpha +2 = 0


x=\dfrac{4\cos \alpha \pm\sqrt{16\cos^2\alpha -8 } }{4}

=\cos \alpha \pm \dfrac{1}{2} \sqrt{4\cos^2 \alpha -2 }
(1)

x^2-4x\sin \alpha -2=0

x=\dfrac{4\sin \alpha \pm\sqrt{16\sin^2\alpha +8 } }{-4}

=-\sin \alpha \pm \dfrac{1}{2}\sqrt{4\sin^2 \alpha +2 }
(2)


Since the radicand of the radical of (1) is larger or equal to zero,

16\cos^2 \alpha -8 \geq 0

\cos^2 \alpha \geq \dfrac{1}{2}

\alpha \leq \dfrac{\pi}{6}


Therefore, cancel the negative case of (1) and (2) and yield the following equation.

\cos \alpha + \dfrac{1}{2} \sqrt{4\cos^2 \alpha -2 } = -\sin \alpha + \dfrac{1}{2}\sqrt{4\sin^2 \alpha +2 }

2(\sin \alpha +\cos \alpha ) = \sqrt{4\sin^2 \alpha +2 }- \sqrt{4\cos^2 \alpha -2 }


4(1+2\sin \alpha \cos \alpha )

= 4\sin ^2 \alpha +2 + 4\cos ^2 \alpha -2 - 4 \sqrt{ 2\sin^2 \alpha+1}\sqrt{2 \cos^2 \alpha-1}


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

4\sin^2 \alpha \cos^2 \alpha = (2\sin^2 \alpha+1)(2 \cos^2 \alpha-1)

= 4\sin^2 \alpha \cos^2 \alpha - 2\sin^2 \alpha + 2 \cos^2 \alpha -1

Therefore,

\cos^2 \alpha - \sin^2 \alpha = \dfrac{1}{2}

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


2\alpha = \dfrac{\pi}{3}


\alpha = \dfrac{\pi}{6}





Steven Zheng posted 5 months ago

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