# Substitution? Confused!

• Jan 9th 2010, 05:32 AM
Punch
Substitution? Confused!
Gievn that tan x = cot(x) - 2cot(2x),

Show that 2tan20 + 4tan40 + 8tan80 = 9(cot10 - tan10)
• Jan 9th 2010, 05:43 AM
Prove It
Quote:

Originally Posted by Punch
Gievn that tan x = cot(x) - 2cot(2x),

Show that 2tan20 + 4tan40 + 8tan80 = 9(cot10 - tan10)

$\displaystyle 2\tan{20} + 4\tan{40} + 8\tan{80}$

$\displaystyle = 2(\cot{20} - 2\cot{40}) + 4(\cot{40} - 2\cot{80}) + 8(\cot{80} - 2\cot{160})$

$\displaystyle = 2\cot{20} - 4\cot{40} + 4\cot{40} - 8\cot{80} + 8\cot{80} - 16\cot{160}$

$\displaystyle = 2\cot{20} - 16\cot{160}$

$\displaystyle = 2\cot{20} - \frac{16}{\tan{160}}$.

Since $\displaystyle 160^\circ$ is in the second quadrant, $\displaystyle \tan{160} = -\tan{20}$.

Continuing...

$\displaystyle = 2\cot{20} + \frac{16}{\tan{20}}$

$\displaystyle = 2\cot{20} + 16\cot{20}$

$\displaystyle = 18\cot{20}$.

Now we use the double angle identity for cotangent.

$\displaystyle \cot{2\theta} = \frac{\cot^2{\theta} - 1}{2\cot{\theta}}$.

So $\displaystyle \cot{20} = \cot{(2\cdot 10)} = \frac{\cot^2{10} - 1}{2\cot{10}}$.

Continuing...

$\displaystyle = 18\left(\frac{\cot^2{10} - 1}{2\cot{10}}\right)$

$\displaystyle = 18\left(\frac{1}{2}\cot{10} - \frac{1}{2}\tan{10}\right)$

$\displaystyle = 9(\cot{10} - \tan{10})$.
• Jan 9th 2010, 04:28 PM
Punch
why do i have to change $\displaystyle [2\cot{20} - \frac{16}{\tan{160}}]$ to $\displaystyle [2\cot{20} + \frac{16}{\tan{20}}]$, is it possible work with the tan120 below?

if it's not possible, then may i would like to ask if it is true that i have to change all the angles to acute?
• Jan 9th 2010, 05:31 PM
Prove It
Quote:

Originally Posted by Punch
why do i have to change $\displaystyle [2\cot{20} - \frac{16}{\tan{160}}]$ to $\displaystyle [2\cot{20} + \frac{16}{\tan{20}}]$, is it possible work with the tan120 below?

if it's not possible, then may i would like to ask if it is true that i have to change all the angles to acute?

Yes you have to change all the angles to acute.

It would make sense, since you're trying to get all the angles to $\displaystyle 10^\circ$...
• Jan 10th 2010, 06:26 AM
Once you get to $\displaystyle 18\cot20$, use the original result, re-arranged as:
$\displaystyle 2\cot2x= \cot x -\tan x$
$\displaystyle \Rightarrow 18\cot20=9(\cot10 -\tan10)$