# relation between tan theta and tan 2theta

• April 20th 2006, 07:41 AM
ling_c_0202
relation between tan theta and tan 2theta
I have got a question on the relation between [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed]

The question consists of two parts:
The first pary asks me to solve the equation [LaTeX ERROR: Convert failed]
Then I 've got the correct ans, which is

[-1 + (1+a^2)^1/2]
_________________
a

or

[-1 - (1+a^2)^1/2]
_________________
a

Then the second part ask me to express [LaTeX ERROR: Convert failed] in terms of [LaTeX ERROR: Convert failed] and the ans is

[-1 + (1+ (tan 2 theta) ^2 )^1/2]
___________________________________
a

or

[-1 - (1+ (tan 2 theta) ^2 )^1/2]
___________________________________
a
• April 20th 2006, 08:29 AM
CaptainBlack
$
\tan(2\theta)=\frac{\sin(2 \theta)}{\cos(2 \theta)}=\frac{2\sin(\theta)\cos(\theta)}{(\cos( \theta ))^2-(\sin(\theta))^2}
$

So:

$
\frac{1}{\tan(2\theta)}=\frac{1}{\tan(\theta)}-\tan(\theta)
$

then multiplying through by $\tan(\theta)$ and rearranging gives:

$
(tan(\theta))^2+\frac{1}{\tan(2 \theta)}\tan(\theta)-1=0
$

from which (if the algebra has been done right) the required result should
follow from the application of the previous result or the quadratic formula.

RonL