# inverse of trigonometry

• December 19th 2011, 09:55 PM
SiddS
inverse of trigonometry
How did 15 degree come in the following equation.

plz help.

• December 19th 2011, 11:41 PM
CaptainBlack
Re: inverse of trigonometry
Quote:

Originally Posted by SiddS
How did 15 degree come in the following equation.

plz help.

1. It is not 15 but -15 degrees.

2. Because this involves $\sqrt{3}$ you need to think in terms of 30 or 60 degrees, so look up the tangent half angle formula.

CB
• December 21st 2011, 07:31 AM
sbhatnagar
Re: inverse of trigonometry
$\theta = \tan^{-1}{\left( \frac{1-\sqrt{3}}{1+\sqrt{3}}\right)}=\tan^{-1}{\left( \frac{1-\sqrt{3}}{1+(1)(\sqrt{3})}\right)}$

Using the formula: $\tan^{-1}(x)-\tan^{-1}(y)=\tan^{-1}{\left( \frac{x-y}{1+xy}\right)}$

we get $\theta= \tan^{-1}(1)-\tan^{-1}(\sqrt{3})= 45^{\circ}-60^{\circ}=-15^{\circ}$