# tan(15)?

• October 22nd 2006, 09:25 PM
chancey
tan(15)?
I am trying to work out the exact value of tan(15) using the ratios in terms of tan. But the book doesn't make any sense. Can some please explain how to solve this?
• October 22nd 2006, 10:59 PM
Soroban
Hello, chancey!

Quote:

Find the exact value of $\tan 15^o$

You're expected to know the double-angle identities:

. . $\sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{2}} \qquad \cos\frac{\theta}{2} = \sqrt{\frac{1 + \cos\theta}{2}} \qquad \tan\frac{\theta}{2}\;=\;\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}$

Let $\theta = 30^o$
We have: . $\tan15^o \;= \;\sqrt{\frac{1-\cos30^o}{1 + \cos30^o}} \;= \;\sqrt{\frac{1 - \frac{\sqrt{3}}{2}} {1 + \frac{\sqrt{3}}{2}}} \;= \;\sqrt{\frac{2 - \sqrt{3}} {2 + \sqrt{3}}}$

Rationalize: . $\sqrt{\frac{2 - \sqrt{3}}{2 + \sqrt{3}}\cdot\frac{2 - \sqrt{3}}{2 - \sqrt{3}}} \;=\;\sqrt{\frac{(2 - \sqrt{3})^2}{4 - 3}} \;=\;\boxed{2 - \sqrt{3}}$

• October 22nd 2006, 11:00 PM
CaptainBlack
Quote:

Originally Posted by chancey
I am trying to work out the exact value of tan(15) using the ratios in terms of tan. But the book doesn't make any sense. Can some please explain how to solve this?

Start with $\tan(30) = 1/\sqrt{3}$, then use:

$
\tan(2A)=\frac{2 \tan(A)}{1+\tan^2(A)}
$

Put $A=15$ then you have:

$
\tan(30)=\frac{2 \tan(15)}{1+\tan^2(15)}=\frac{1}{\sqrt{3}}
$

Which is a quadratic in $\tan(15)$, ie put $u=\tan(15)$ then:

$
\frac{2u}{1-u^2}=1/\sqrt{3}
$

or:

$
u^2 + 2\sqrt{3}u-1=0
$

Which has roots $u=\pm 2-\sqrt{3}$, of these we want the positive root so $\tan(15)=2-\sqrt{3}$.

RonL