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complex2 q (1 Viewer)
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yeah the roots would be
lets say you start off with
and then
the product of those pairs of roots will cancel each other out when adding them to each other.
so basically you get all but three terms cancelling out, which is where the root multiplies by their negative equivalent ie
alternatively i think you might be able to do the question by letting
hogzillaAnson
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Sums of pairs would work but there's an easier way by using sin^2 + cos^2 = 1:
^2 &= \cos ^2 \frac{\pi}{12} + \cos^2 \left(\frac{5\pi}{12} \right) + 2\cos \frac{\pi}{12} \cos \frac{5\pi}{12} \\ &= \cos^2 \frac{\pi}{12} + \cos^2 \left( \frac{\pi}{2} - \frac{\pi}{12}\right) + 2\cdot \frac{1}{4}\ \ [\text{Using previous result}]\\ &= \cos^2 \frac{\pi}{12} + \sin^2 \frac{\pi}{12} + \frac{1}{2}\\ &= 1+ \frac{1}{2} = \frac{3}{2}\ \ [\text{Fundamental trigonometric identity}] \\ \left( \cos \frac{\pi}{12} + \cos \frac{5\pi}{12} \right) &= \sqrt{\frac{3}{2}}\ \ \text{Noting that } \cos\frac{\pi}{12} + \cos \frac{5\pi}{12} > 0\ \text{since}\ \cos\theta\ \text{is decreasing on the interval}\ \theta\in(0,\pi/2]\ \text{and}\ \cos \frac{\pi}{2} = 0\\ \text{The identity}\sin\theta &= \cos \left(\frac{\pi}{2} - \theta\right)\ \text{is pretty handy to know both for Ext 1 and Ext 2}\end{align*})
From this point, part (d) reduces to solving a quadratic equation.
 x + \alpha \beta = 0 \\ 0 &= x^2 - \sqrt{\frac{3}{2}}x + \frac{1}{4}\\ x &= \frac{\sqrt{3/2} \pm \sqrt{3/2 - 4\cdot 1/4}}{2} = \frac{\sqrt{3/2} \pm \sqrt{1/2}}{2} \\ &= \frac{\sqrt{3} \pm 1 }{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6} \pm \sqrt{2}}{4} \\ \text{Since}\ \alpha > \beta, \cos \frac{\pi}{12} &= \frac{\sqrt{6}+\sqrt{2}}{4}, \cos \frac{5\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4} \end{align*})
From this point, part (d) reduces to solving a quadratic equation.
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