Question

Which identity would I use to find the exact value of tan(pi/8)

Answer 1

tanx=2tan(x/2)/(1-tan^2(x/2))

Answer 2

where x=pi/4

Answer 3

But wouldn't that be solving for x=pi/4 instead of x=pi/8 ? And what is the name of that identity, I can't find it on my identities sheet

Answer 4

nah, we know the value of tanpi/4, so we will get a quadratic for tanpi/8, whose 1 root is positive nd other negative, we will reject negative by seeing the quadrant

Answer 5

this is called half angle identity

Answer 6

how about\[\tan(a\pm b)={\tan a\pm\tan b\over1\mp\tan a\tan b}\]??

Answer 7

sir turning test, what will u put a nd b?

Answer 8

\[
1=\frac{2 u}{1-u^2}
\]
Solve the above for \[ \tan (\pi/4)

Answer 9

Solve the above for \[ \tan (\pi/4)\]

Answer 10

I'm no sir :)
I was just brainstorming...

Answer 11

I guess you guys are right, news to me

Answer 12

elias: Would I plug in pi/4 for u?

Answer 13

ok sorry sir, ( it is out of habit, i cant help calling someone sir or mam,) u nd i both will have to live with it

Answer 14

In solving for u you get
\[

\begin{array}{c}
u=-1-\sqrt{2} \\
u=-1+\sqrt{2} \\
\end{array}

\]
Which one do you choose?

Answer 15

If you put \( x = \frac \pi 4\) in
\[
\tan (x)=\frac{2 \tan
\left(\frac{x}{2}\right)}{1-\tan
^2\left(\frac{x}{2}\right)}
\]

Answer 16

put now \( u =\tan(\frac x 2) \) you get
\[
1=\frac{2 u}{1-u^2}
\]
and you finish as above.

Answer 17

\[
1-u^2=2u\\
u^2 + 2 u -1 =0
\]

Answer 18

Solving it you get
\[

\begin{array}{c}
u=-1-\sqrt{2} \\
u=-1+\sqrt{2} \\
\end{array}

\]

You choose the seconf one. Why?

Answer 19

\[
u=\tan(\pi/8)=-1+\sqrt{2}
\]

Answer 20

Ok, I'm going to work through this one, thanks for the explanation elias

Answer 21

@TuringTest I am goin with him