Inverse trigonometric function

**pankaj** · July 5th 2009, 07:37 PM

If $A=\tan^{-1}(2\sqrt2-1)$ and $B=3\sin^{-1}\left(\frac{1}{3}\right)+\sin^{-1}\left(\frac{3}{5}\right)$

Without using calculator prove that $A>B$

**apcalculus** · July 5th 2009, 08:37 PM

Originally Posted by pankaj

If $A=\tan^{-1}(2\sqrt2-1)$ and $B=3\sin^{-1}\left(\frac{1}{3}\right)+\sin^{-1}\left(\frac{3}{5}\right)$

Without using calculator prove that $A>B$

$\tan(A) = 2\sqrt{2} - 1$

$M = 3 \sin^{-1}\frac{1}{3}$
$N = sin^{-1}\frac{3}{5}$
$\sin \frac{M}{3} = \frac{1}{3}$
$\sin N = \frac{3}{5}$
$B=M + N$
$\tan B = \tan (M + N)=\frac{tanM + tanN}{1 - tanM tanN}$

$\tan N = \frac{\sin N}{\sqrt{1- \sin^2 N}}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$

$\tan M = \frac{\sin M}{\sqrt{1- \sin^2 M}}=?$

**Grandad** · July 5th 2009, 10:00 PM

Hello pankaj

Originally Posted by pankaj

If $A=\tan^{-1}(2\sqrt2-1)$ and $B=3\sin^{-1}\left(\frac{1}{3}\right)+\sin^{-1}\left(\frac{3}{5}\right)$

Without using calculator prove that $A>B$

First, let's note that A and B can have infinitely many values. Some will give A > B and some will give A < B. So I think we must assume that all inverse trig functions are assumed to be principle values.

Secondly, on this assumption, I'm afraid that it just isn't true! Try it with your calculator, and you'll see.

Could you please check that we have the correct question here?

Grandad

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