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2 π sin 1 ( 240 289 ) cos 1 ( 15 17 ) = ? \frac{2\pi-\sin^{-1}\left(\frac{240}{289}\right)}{\cos^{-1}\left(-\frac{15}{17}\right)}=\ ?


The answer is 2.

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1 solution

Pi Han Goh
Mar 28, 2021

We start with a 8-15-17 right-triangle. Let one of its angles be θ \theta such that cos θ = 15 17 \cos\theta = \frac{15}{17} .

By double angle formula , cos ( 2 θ ) = 2 cos 2 θ 1 = 2 ( 15 17 ) 2 1 = 161 289 \cos(2\theta) = 2\cos^2\theta - 1 = 2(\tfrac{15}{17})^2 - 1 = \frac{161}{289} .

We can then construct another right-triangle of side lengths 161-240-289, and with an angle 2 θ 2\theta for which cos ( 2 θ ) = 161 289 \cos(2\theta) = \frac{161}{289} or sin ( 2 θ ) = 240 289 \sin(2\theta) = \frac{240}{289} . Then,

2 θ = 2 θ 0 sin 1 ( 240 289 ) = 2 cos 1 ( 15 17 ) 0 2 π sin 1 ( 240 289 ) = 2 π 2 cos 1 ( 15 17 ) 0 = 2 [ π cos 1 ( 15 17 ) ] 0 = 2 cos 1 ( 15 17 ) 0 2 π sin 1 ( 240 289 ) cos 1 ( 15 17 ) = 2 \begin{array} { r c l} 2\theta &=& 2 \cdot \theta \\ \phantom0 \\ \sin^{-1} \left( \dfrac{240}{289} \right) &=& 2 \cdot \cos^{-1} \left ( \dfrac{15}{17} \right) \\ \phantom 0 \\ 2\pi - \sin^{-1} \left( \dfrac{240}{289} \right) &=& 2\pi - 2 \cos^{-1} \left ( \dfrac{15}{17} \right) \\ \phantom 0 \\ &=& 2 \left [ \, \pi - \cos^{-1} \left ( \dfrac{15}{17} \right) \, \right ] \\ \phantom 0 \\ &=& 2 \cos^{-1} \left ( -\dfrac{15}{17} \right) \\ \phantom 0 \\ \dfrac{2\pi - \sin^{-1} \left( \dfrac{240}{289} \right)}{\cos^{-1} \left ( -\dfrac{15}{17} \right)} &=& \boxed2 \end{array}

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