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If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\displaystyle \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$

$\bf{My\; Try::}$ Given $$ \alpha = \arctan \bigg(\frac{2(2\sqrt{2}-1)}{1-(2\sqrt{2}-1)^2}\bigg)=\arctan \bigg(\frac{(2\sqrt{2}-1)}{2\sqrt{2}-4}\bigg)$$

and $$\beta = \arcsin\bigg(1-\frac{4}{27}\bigg)+\arcsin\bigg(\frac{3}{5}\bigg) = \arcsin\bigg(\frac{23}{27}\bigg)+\arcsin\bigg(\frac{3}{5}\bigg)$$

So $$\arcsin\bigg[\frac{23}{27}\cdot \frac{4}{5}+\frac{3}{5}\cdot \frac{10\sqrt{2}}{23}\bigg]$$

Now how can i solve it, Help required, Thanks

juantheron
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3 Answers3

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HINT:

As $\arcsin\dfrac13<\arcsin\dfrac12=\dfrac\pi6$

$3\arcsin\dfrac13=\arcsin\dfrac{23}{27}=\arctan\dfrac{23}{10\sqrt2}$

$\arcsin\dfrac35=\arctan\dfrac34$

As $\dfrac{23}{10\sqrt2}\cdot\dfrac34>1,$

using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,

$$\arctan\dfrac{23}{10\sqrt2}+\arctan\dfrac34=\pi+\arctan\dfrac{\dfrac{23}{10\sqrt2}+\dfrac34}{1-\dfrac{23}{10\sqrt2}\cdot\dfrac34}$$

As $2\sqrt2-1>1$ $$2\arctan(2\sqrt{2}-1)=\pi+\arctan ?$$

Now for $\dfrac\pi2>a>b>-\dfrac\pi2,\arctan a>\arctan b $

Community
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lab bhattacharjee
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On the one hand we have $$\tan{\alpha\over2}=2\sqrt{2}-1>\sqrt{3}=\tan{\pi\over3}\ ,$$ hence ${\alpha\over2}>{\pi\over3}$, or $$\alpha>{2\pi\over3}\ .$$ On the other hand the convexity of $$t\mapsto\arcsin t\qquad(0\leq t\leq{\pi\over2})$$ implies

$$\arcsin{1\over3}<{2\over3}\arcsin{1\over2}={2\over3}\>{\pi\over6}={\pi\over9}\ .$$ Furthermore $\arcsin{3\over5}<\arcsin{1\over\sqrt{2}}={\pi\over4}$. It follows that $$\beta<3{\pi\over9}+{\pi\over4}={7\pi\over12}<\alpha\ .$$

Christian Blatter
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\begin{align} \beta&=\sqrt{10}\left(\frac{3}{\sqrt{10}}\arcsin\frac13+\frac{1}{\sqrt{10}}\arcsin\frac35\right)\\ &<\sqrt{10}\arcsin\left(\frac{4\sqrt{10}}{25}\right) \end{align} where I use the convexity of $\arcsin$. On the other hand using the definition of $\arctan$ we obtain $$\arcsin\left(\frac{4\sqrt{10}}{25}\right)=\arctan\left(\frac{4\sqrt{2}}{\sqrt{93}}\right).$$ Now further note that $\sqrt{10}<4$ hence we should check wether $$2\arctan\left(\frac{2\sqrt{2}}{\sqrt{93}}\right)<\arctan(2\sqrt2-1)$$ We hence (with $\tan2x=\frac{2\tan x}{1-\tan^2x}$) should prove that $$\frac{8\sqrt{186}}{61}<2\sqrt2-1.$$

Math-fun
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