Simple Trig Question

Question

Hi I am not sure I am solving this trig question correctly:

$\tan\left(\sin^{-1}\left(\dfrac{2\sqrt{x}}{1+x}\right)\right) = ?$

I drew a right triangle and set an angle equal to $\sin^{-1}\left(\dfrac{2\sqrt{x}}{1+x}\right)$

so that I could find the third (adjacent) side after setting the opposite as $2\sqrt{x}$ and the hypotenuse as $1+x$ by using the Pythagorean theorem.

I believe after using the Pythagorean theorem and doing the algebra that the adjacent side is $\sqrt{x^2-2x+1}$

So would it then follow that the answer to the question would be $\dfrac{2\sqrt{x}}{\sqrt{x^2-2x+1}}$?

thanks

Alternatively, you can rewrite $\tan;x$ as $\frac{\sin;x}{\sqrt{1-\sin^2 x}}$... — J. M. ain't a mathematician, May 05 '11 at 23:58

score 4 · Accepted Answer · answered May 05 '11 at 23:54

4

Yes, you did good. The denominator can be simplified to $|x-1|$.

answered May 05 '11 at 23:54

lhf

216,483

It is a multiple choice question and the answer I got wasn't an option but $\dfrac{2\sqrt{x}}{x-1}$ is. how can it be simplified? by rationailizing? – Matt May 05 '11 at 23:58
1

Look at the denominator of your answer. Look at the denominator in the option. What do you see? – Gerry Myerson May 06 '11 at 00:00
ahhh, $(x-1)^2$, thanks! – Matt May 06 '11 at 00:07
1

@Matt: Take, for example, $x=1/4$. Then your answer is right but the one that comes from $\frac{2\sqrt{x}}{x-1}$ is wrong. – André Nicolas May 06 '11 at 00:23

score 3 · Answer 2 · 2011-05-08T14:57:20.933

3

Another way of doing this by substitution. Note that $\sin{2x} =\frac{2 \tan{x}}{1+\tan^{2}{x}}$. So what I did is to put $x = \tan^{2}{t}$. Then what you want is precisely $\tan{2t}$. Which you can obtain by the formula \begin{align*} \tan{2t} &=\frac{2 \cdot \tan{t}}{1 - \tan^{2}{t}} \\ &= \frac{2 \sqrt{x}}{1-x} \qquad \Bigl[ \text{since} \ x=\tan^{2}{t} \ \Bigr] \end{align*}

edited May 08 '11 at 14:57

answered May 07 '11 at 11:24

@Matt: Hope this solution helps. – May 08 '11 at 14:56

Simple Trig Question

2 Answers2