• Dec 8th 2008, 06:13 PM
lanvin
we took down a note, where the example showed us how to find the exact value of

tan (-5π / 12)

then she wrote down....

tan (-π/4) - tan (π/6)
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
1+tan(-π/4)tan(π/6)

and then from that, she jumped to the final answer, which is...

(-√3-1) / (-√3-1)

How did she do that?
Maybe I copied down her note wrong, cuz I keep getting a different answer...
• Dec 8th 2008, 06:25 PM
11rdc11
Quote:

Originally Posted by lanvin
we took down a note, where the example showed us how to find the exact value of

tan (-5π / 12)

then she wrote down....

tan (-π/4) - tan (π/6)
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ
1+tan(-π/4)tan(π/6)

and then from that, she jumped to the final answer, which is...

(-√3-1) / (-√3-1)

How did she do that?
Maybe I copied down her note wrong, cuz I keep getting a different answer...

$\displaystyle \tan{\frac{-5\pi}{12}} = \tan{225}$

Using a reference angle we see that it is an angle of 75 degrees

So now sum difference formula for tan

$\displaystyle \frac{\tan{45}+\tan{30}}{1-\tan{45}\tan{30}}$

$\displaystyle \frac{(1) + (\frac{\sqrt{3}}{3})}{1 - (1)(\frac{\sqrt{3}}{3})}$

Then multiply by 3 to numerator and denominator

$\displaystyle \frac{3 +\sqrt{3}}{3- \sqrt{3}}$

then rationalize

$\displaystyle \frac{6\sqrt{3} + 12}{6}$

$\displaystyle 2 + \sqrt{3}$

Now multiply a neg in because it is in the 4th quad

$\displaystyle -2 - \sqrt{3}$