Trigonometry #106

The following approach is not ideal, but it is all I could come up with.

Note first that $(2\sqrt{2} - 1) \gt \sqrt{3}.$ Thus $A \gt 2\tan^{-1}(\sqrt{3}) = 120^{\circ}.$

Next, note that $\frac{3}{5} \lt \frac{\sqrt{2}}{2}$ , and so $\sin^{-1}(\frac{3}{5}) \lt \sin^{-1}(\frac{\sqrt{2}}{2}) = 45^{\circ}.$

Now for the tricky one. We have that

$\sin^{2}(22.5^{\circ}) = \dfrac{1}{2}(1 - \sin(45^{\circ})) = \dfrac{2 - \sqrt{2}}{4} \gt \dfrac{\frac{1}{2}}{4} = \dfrac{1}{8} \gt \dfrac{1}{9} = (\dfrac{1}{3})^{2}.$

This implies that $\sin^{-1}(\frac{1}{3}) \lt 22.5^{\circ}$ , and thus that

$B \lt (3*22.5^{\circ} + 45^{\circ}) = 112.5^{\circ} \lt A.$

The greater of the two angles is therefore $\boxed{A}$ .

$2\sqrt{2} - 1 > 0 \Rightarrow 0 < \dfrac{A}{2} < \dfrac{\pi}{2} \Rightarrow 0 < A < \pi$

$\tan\left(\dfrac{A}{2}\right) = 2\sqrt{2} - 1 \Rightarrow \cos\left(\dfrac{A}{2}\right) = \dfrac{1}{\sqrt{10-4\sqrt{2}}} \Rightarrow \cos(A) = \dfrac{2\sqrt{2}-4}{5-2\sqrt{2}}$

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$0 < \sin^{-1}\left(\dfrac13\right) < \sin^{-1}\left(\dfrac12\right) = \dfrac{\pi}{6} \Rightarrow 0 < B < 3\times{\dfrac{\pi}{6}} + \dfrac{\pi}{2} = \pi$

$\sin^{-1}\left(\dfrac{1}{3}\right) = \theta , \sin^{-1}\left(\dfrac35\right) = \beta$

$\cos(3\theta + \beta) = \cos(B) = \dfrac{40\sqrt{2}\pm69}{135}$

$3\theta \in{(0 , \pi)} \Rightarrow \sin(3\theta) > 0 \Rightarrow \cos(B) = \dfrac{40\sqrt{2}-69}{135}$

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$\cos(B) > \cos(A) \Rightarrow A > B$