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Geometry Level 4

If tan ⁡ ( α − β ) = sin ⁡ ( 2 β ) 3 − cos ⁡ ( 2 β ) \large{\tan(\alpha-\beta)=\frac{\sin(2\beta)}{3-\cos(2\beta)}} then tan ⁡ ( α ) = f ( β ) \tan(\alpha)=f(\beta) . Find f ( π 3 ) \large{f(\frac{\pi}{3})} .


The answer is 3.46.

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Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

For β = π 3 \beta=\frac{\pi}{3} , then:

tan ⁡ ( α − π 3 ) = sin ⁡ 2 π 3 3 − cos ⁡ 2 π 3 tan ⁡ α − tan ⁡ π 3 1 + tan ⁡ α tan ⁡ π 3 = 3 2 3 − ( − 1 2 ) tan ⁡ α − 3 1 + 3 tan ⁡ α = 3 7 7 tan ⁡ α − 7 3 = 3 + 3 tan ⁡ α 4 tan ⁡ α = 8 3 tan ⁡ α = f ( π 3 ) = 2 3 = 3.46 \begin{aligned} \tan{\left(\alpha-\frac{\pi}{3}\right)} & = \dfrac{\sin{\frac{2\pi}{3}}}{3-\cos{\frac{2\pi}{3}}} \\ \frac{\tan{\alpha}-\tan{\frac{\pi} {3}}}{1+\tan{\alpha} \tan{\frac{\pi} {3}}} & = \frac{\frac{\sqrt{3}}{2}}{3-\left(-\frac{1}{2}\right)} \\ \frac{\tan{\alpha}-\sqrt{3}}{1+ \sqrt{3} \tan{\alpha}} & = \frac{\sqrt{3}}{7} \\ 7\tan{\alpha}-7\sqrt{3} & = \sqrt{3} + 3 \tan{\alpha} \\ 4\tan{\alpha} & = 8\sqrt{3} \\ \tan{\alpha} & = f\left(\frac{\pi}{3} \right) = 2\sqrt{3} = \boxed{3.46} \end{aligned}

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