tan 70 = tan 20 + 2 tan 40 +4 tan 10

**perash** · June 27th 2007, 06:35 AM

prove

tan 70 = tan 20 + 2 tan 40 +4 tan 10

**red_dog** · June 27th 2007, 07:37 AM

We have the identity $tan x -cotan x=-2cotan 2x$ .
Then we have
$tan20-cotan20=-2cotan40$
$2tan40-2cotan40=-4cotan80$
$4tan10=4cotan80$
Adding the three equalities it results
$tan20+2tan40+4tan10=cotan20=tan70$

**Soroban** · June 27th 2007, 08:42 AM

Hello, perash!

red_dog had a brilliant solution!
. . Did you follow it?

Prove: . $\tan70 \;= \;\tan20 + 2\tan40 + 4\tan10$

We have the identity: . $\tan\theta - \cot\theta \:=\:-2\cot2\theta$ .**

So we have: . $\begin{array}{ccc}\tan20 - \cot20 & = & -2\cot40 \\ 2(\tan40 - \cot40) & = & -4\cot80 \\ 4\tan10 & = & 4\cot 80 \end{array}$

Add the three equations:
. . $\tan20 -\cot20 + 2\tan40 - 2\cot40 + 4\tan10 \;=\;-2\cot40 - 4\cot80 + 4\cot80$

which simplifies to:
. . $\tan20 - \cot20 + 2\tan40 + 4\tan10 \:=\:0$

. . . . . . . $\tan20 + 2\tan40 + 4\tan10 \;=\;\cot20$

. . . . . . . $\tan20 + 2\tan40 + 4\tan10 \;=\;\tan70$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
We have: . $\tan\theta - \cot\theta \;=\;\tan\theta - \frac{1}{\tan\theta} \;=\;\frac{\tan^2\theta - 1}{\tan\theta} \;=\;\frac{-(1 - \tan^2\theta)}{\tan\theta}$

. . $= \;\frac{-2(1-\tan^2\theta)}{2\tan\theta} \;=\;\frac{-2}{\frac{2\tan\theta}{1-\tan^2\theta}} \;=\;\frac{-2}{\tan2\theta} \;=\;-2\cot2\theta$

. . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Alternate proof:

$\tan\theta - \cot\theta \;=\;\frac{\sin\theta}{\cos\theta} - \frac{\cos\theta}{\sin\theta} \;= \;\frac{\sin^2\theta - \cos^2\theta}{\sin\theta\cos\theta} \;=\;\frac{-(\cos^2\theta - \sin^2\theta)}{\sin\theta\cos\theta}$

. . $= \;\frac{-2(\cos^2\theta - \sin^2\theta)}{2\sin\theta\cos\theta} \;=\;\frac{-2\cos2\theta}{\sin2\theta} \;=\;-2\cot2\theta$

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