Hardest Inverse and Binomial questions? (2 Viewers)

SammyT123 · May 25, 2016

Hey all,

If any of you remember any weird, tricky, interesting or challenging questions from Inverse(functions/trig) and Binomial theorem, can you please post them here

Fanks

Trebla · May 25, 2016

See BoS trials

kthanxbye

Paradoxica · May 25, 2016

Drongoski · May 25, 2016

Paradoxica said:
$$\noindent Show/prove, Geometrically, that $\sin^{-1}{\frac{3}{5}}+\sin^{-1}{\frac{12}{13}}+\sin^{-1}{\frac{63}{65}}=\pi \\ $Show/prove, for $|t| \le 1$, $\sin^{-1}{\left(\frac{2t}{1+t^2}\right)} \equiv 2\tan^{-1}{t} \\ $Hence, or otherwise, show/prove $\tan^{-1}{\frac{1}{3}} + \tan^{-1}{\frac{2}{3}} + \tan^{-1}{\frac{7}{9}} = \frac{\pi}{2}$

Q1: The 3 Pythagorean triples are: {3,4,5}, {5,12,13} and {16,63,65} = {3.2,13,12.6}

You can draw the 3 right-angled triangles associate with these triples: {3,4,5}, {5,12,13} and {3.2,12.6,13}

You draw the 1st triangle with side 4 horizontal. You then erect the 2nd right-angled triangle with the 5 side resting on the hypotenuse of the 1st. You finally allign the 13 (hypotenuse side) of the 3rd triangle withe the 13 (hypotenuse) side of the 2nd triangle. Then there you have it: all the 3 adjacent acute angles:

$sin^{-1}\frac {3}{5}, $ $ sin^{-1}\frac {12}{13} $ and $ sin^{-1}\frac {63}{65}$

forming a straight angle.

Of course a simple diagram would have made this easier to see; but I just can't do it. Some kind soul can easily do this and display the diagram.

Drongoski · May 26, 2016

Q2

The 1st part is the usual t-formulae

Now applying this result (omitting much detail because of LaTeX tedium):

$2tan^{-1}\frac {1}{3} = sin^{-1 \left(\frac {2 \times \frac{1}{3}}{1+(\frac {1}{3})^2} \right )} = sin^{-1}\frac{3}{5}\\ \\ 2tan^{-1}\frac {2}{3} = sin^{-1} \frac {12}{13}\\ \\ 2tan^{-1}\frac {7}{9} = sin^{-1}\frac {63}{65}\\ \\ \therefore 2(tan^{-1}\frac {1}{3}+tan^{-1}\frac{2}{3}+tan^{-1}\frac {7}{9}) \\ \\ = sin^{-1}\frac{3}{5} + sin^{-1}\frac {12}{13} + sin^{-1}\frac {63}{65}= \pi$

by virtue of Q1 result.

Hence the result.

Alternatively, by brute force:

By repeated application of the formula:

$tan(\alpha + \beta) = \frac {tan \alpha + tan \beta}{1 - tan\alpha \tan \beta}\\ \\ $you will end up with $ tan($sum of the 3 inverse tangents $) \\ \\ = \frac {\frac {1}{3} + 3}{1 - \frac{1}{3} \times 3} = \frac {\frac {10}{3}}{1 - 1} = \infty$

So that angle must be pi/2

Hardest Inverse and Binomial questions? (2 Viewers)

SammyT123

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Trebla

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Paradoxica

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Drongoski

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Drongoski

Well-Known Member

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