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Algebra Difficulty 5.5 AIME, harder Find the answer

3.193.

2cos⁡(π6−2α)−3sin⁡(5π2−2α)cos⁡(9π2−2α)+2cos⁡(π6+2α)=tan⁡2α3 \frac{2 \cos \left(\frac{\pi}{6}-2 \alpha\right)-\sqrt{3} \sin \left(\frac{5 \pi}{2}-2 \alpha\right)}{\cos \left(\frac{9 \pi}{2}-2 \alpha\right)+2 \cos \left(\frac{\pi}{6}+2 \alpha\right)}=\frac{\tan 2 \alpha}{\sqrt{3}}

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

## Solution.

2cos⁡(π6−2α)−3sin⁡(5π2−2α)cos⁡(9π2−2α)+2cos⁡(π6+2α)==2(cos⁡π6cos⁡2α+sin⁡π6sin⁡2α)−3cos⁡2αsin⁡2α+2(cos⁡π6cos⁡2α−sin⁡π6sin⁡2α)==2(32cos⁡2α+12sin⁡2α)−3cos⁡2αsin⁡2α+2(32cos⁡2α−12sin⁡2α)= \begin{aligned} & \frac{2 \cos \left(\frac{\pi}{6}-2 \alpha\right)-\sqrt{3} \sin \left(\frac{5 \pi}{2}-2 \alpha\right)}{\cos \left(\frac{9 \pi}{2}-2 \alpha\right)+2 \cos \left(\frac{\pi}{6}+2 \alpha\right)}= \\ = & \frac{2\left(\cos \frac{\pi}{6} \cos 2 \alpha+\sin \frac{\pi}{6} \sin 2 \alpha\right)-\sqrt{3} \cos 2 \alpha}{\sin 2 \alpha+2\left(\cos \frac{\pi}{6} \cos 2 \alpha-\sin \frac{\pi}{6} \sin 2 \alpha\right)}= \\ = & \frac{2\left(\frac{\sqrt{3}}{2} \cos 2 \alpha+\frac{1}{2} \sin 2 \alpha\right)-\sqrt{3} \cos 2 \alpha}{\sin 2 \alpha+2\left(\frac{\sqrt{3}}{2} \cos 2 \alpha-\frac{1}{2} \sin 2 \alpha\right)}= \end{aligned}

=3cos⁡2α+sin⁡2α−3cos⁡2αsin⁡2α+3cos⁡2α−sin⁡2α=sin⁡2α3cos⁡2α=tg⁡2α3. =\frac{\sqrt{3} \cos 2 \alpha+\sin 2 \alpha-\sqrt{3} \cos 2 \alpha}{\sin 2 \alpha+\sqrt{3} \cos 2 \alpha-\sin 2 \alpha}=\frac{\sin 2 \alpha}{\sqrt{3} \cos 2 \alpha}=\frac{\operatorname{tg} 2 \alpha}{\sqrt{3}} .

The identity is proven.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.