Question

Just another cute problem,

Prove that: \( \large \tan \frac \pi {16} + 2 \tan \frac \pi {8} + 4 = \cot \frac{\pi}{16} \)

Enjoy!

Answer 1

If we substitute cos and sin instead of tan and then use the trig identity that\[\sin^{2}x + \cos^{2}x = 1\]we can prove it, I think. But I may have commited a minor mistake in my proof. I will proofread it.

Answer 2

hmm since there is no variable, we are only proving one specific case so verifying rhs equals lhs is sufficient no?

Answer 3

The quickest way seems to be rewritting sin and cos using series expasions and to do long division on them to solve it, haha. :-)

Answer 4

@dumbcow That works! as long as you are deriving RHS wholly from the LHS.

PS: The actual problem came with 4 options.

Answer 5

math is not cute. math is ugly. very ugly.

Answer 6

I derived it from half angles formulas. I think it's an okay proof, but too messy. I would like to see a cleaner and more elegant proof than mine, if possible :-)

Answer 7

My proof is of 3 lines.

Answer 8

I figured it out that my proof is very inelegant. Still, any tips for the cuter solution? :-)

Answer 9

Use the double angle formula for tan to get
\[
\tan(\frac \pi 8)=\sqrt{2}-1
\]
Do the same trick to find
\[\tan(\frac \pi {16})=-1-\sqrt{2}+\sqrt{4+2 \sqrt{2}}
\]
From now it is easy.

I am sure you have an easier proof.

Answer 10

tan (pi/16)-cot(pi/16)+2tan(pi/8)+4
=(2tan(pi/16)/tan(pi/8)) / tan(pi/16) +2 tan(pi/8) +4
=-2cot(pi/8)+2tan(pi/8) +4
=-2(cos pi/8 / sin pi/8 + sinpi/8 /cos pi/8) +4
=-2sin pi/4 / (1/2 sin pi/4) +4
=0

Answer 11

\[\cot (\frac \pi {16}) =1+\sqrt{2}+\sqrt{2 \left(2+\sqrt{2}\right)}
\]
so
\[-1-\sqrt{2}+\sqrt{4+2 \sqrt{2}} + 2 + 2( \sqrt{2}-1) +4=1+\sqrt{2}+\sqrt{2 \left(2+\sqrt{2}\right)}=\cot(\frac \pi{16})
\]

Answer 12

Use the formulas
\[\tan (\frac x 2) =\frac{ 1- \cos(x)}{\sin(x)}\\
\cot (\frac x 2) =\frac{ 1+ \cos(x)}{\sin(x)}\\
\]
Let
\[
x= \frac \pi 8
\]
With this notation we have to prove
\[
\tan(\frac x 2) + 2 \tan(x) - \cot(\frac x 2)=-4\\
\tan(\frac x 2) + 2 \tan(x) - \cot(\frac x 2)=\\
\frac{ 1- \cos(x)}{\sin(x)} + 2 \frac {\sin(x)}{\cos(x)}- \frac{ 1+ \cos(x)}{\sin(x)}=\\
-2 \frac {\cos(x)}{\sin(x)}+ 2 \frac {\sin(x)}{\cos(x)}=\\
\frac {-2\left( \cos^2(x) -\sin^2(x)\right)}
{\sin(x) \cos(x)}=\\

\frac { -2 \cos(2x)}{\frac 12 \sin(2 x)}\\
-4 \frac { \cos(2x)}{ \sin(2 x)}= -4\\
\]
Since
\[
2 x =\frac \pi 4\]

Answer 13

An easier proof. It easy to show that
\[
\tan(v) - \cot (v) =- 2 \cot(2v)

\]
Take
\[ v=\frac \pi{16}
\]
We get
\[\tan(\frac \pi{16}) - \cot (\frac \pi{16}) =- 2 \cot(\frac \pi{8})
\]
What we have to prove becomes
\[ - 2 \cot(\frac \pi{8}) + 2 \tan(\frac \pi{8}) =-4\\

-\cot(\frac \pi{8}) + \tan(\frac \pi{8})=-2\\
\text { By the first formula }\\
- 2 \cot(\frac \pi 4) =-2\\
-2(1) =-2
\]

Answer 14

@bmp: I used the identity \(\tan A = \cot A -2\cot 2A \)

Answer 15

Indeed, that's simpler than what I've done. I did it very much like the first solution posted by @eliassaab.