$\tan(x) = 3$. Find the value of $\sin(x)$ I’m trying to figure out the value for $\sin(x)$ when $\tan(x) = 3$.  The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm doing wrong? 
1.) $\tan(x) = 3$, then $\frac{\sin(x)}{\cos(x)} = 3$. 
2.) Then $\cos(x) = \frac{1}{3}\sin(x)$
3.) $\sin^2 (x) + \left(\frac{1}{3}\sin(x)\right)^2 = 1$    //Pythagorean identity substitution.
4.) $\left(\frac{4}{3}\sin(x)\right)^2 = 1$   //Combining like terms
5.) $\frac{16}{9}\sin^2(x) = 1$   //Square the fraction so I can move it later.
6.) $\sin^2(x) = \frac{1}{\frac{16}{9}}$ //Divide both sides by $\frac{16}{9}$
7.) $\sin^2(x) = \frac{1}{1} * \frac{9}{16} = \frac{9}{16}$  //divide out the fractions
8.) $\sin(x) = \pm \frac{3}{4}$ //square root both sides.
So $\sin(x) = \pm \frac{3}{4}$ but the book says this is wrong. Any help is much appreciated.

In general,
if
$\tan(x) = a$
then
$a^2
=\tan^2(x)
=\dfrac{\sin^2(x)}{\cos^2(x)}
=\dfrac{\sin^2(x)}{1-\sin^2(x)}
$
so
$\dfrac1{a^2}
=\dfrac{1-\sin^2(x)}{\sin^2(x)}
=\dfrac{1}{\sin^2(x)}-1
$
so
$\dfrac{1}{\sin^2(x)}
=\dfrac1{a^2}+1
=\dfrac{1+a^2}{a^2}
$
so
$\sin^2(x)
=\dfrac{a^2}{1+a^2}
$
or
$\sin(x)
=\dfrac{a}{\sqrt{1+a^2}}
$.
Similarly,
$a^2
=\tan^2(x)
=\dfrac{\sin^2(x)}{\cos^2(x)}
=\dfrac{1-\cos^2(x)}{\cos^2(x)}
=\dfrac{1}{\cos^2(x)}-1
$
so
$\dfrac{1}{\cos^2(x)}
=1+a^2
$
so
$\cos(x)
=\dfrac1{\sqrt{1+a^2}}
$.
Now,
try this for
hyperbolic functions.
