Question

Find the value of tan(3pi/16)

Answer 1

.0102812 if you put it into calculator

Answer 2

I need the exact value please... How not sure how... :( I know that we need to use the double angle identities.

Answer 3

hi

Answer 4

Hi freckles! :)

Answer 5

I need some help...

Answer 6

you know 3pi/4 is on the unit circle right?

Answer 7

and we know 1/2*1/2*3pi/4 is 3pi/16

Answer 8

So it sounds like we are going to have to use the lovely half angle identity we used the other day

Answer 9

\[\tan(\frac{\theta}{2})=\frac{\sin(\theta)}{1+\cos(\theta)}\]
have you seen this identity before?

Answer 10

nope...

Answer 11

\[\tan(u)=\frac{\sin(2 u)}{1+\cos(2u)}\]
maybe you seen it written this way
but technically this is the same identity I just mentioned just where theta is 2u

Answer 12

I learned this one:
\[\tan2x=\frac{ 2tanx }{ 1-\tan^2x }\]

Answer 13

ok that can also work

Answer 14

replace 2x with u
so x is u/2

\[\tan(u)=\frac{2 \tan(\frac{u}{2})}{1-\tan^2(\frac{u}{2})}\]

\[u=\frac{3\pi}{4} \\ \tan(\frac{3\pi}{4})=\frac{2 \tan(\frac{ 3 \pi}{8})}{1- \tan^2(\frac{3\pi}{8})}\]
we can evaluate tan(3pi/4) using unit circle

then we want to try to solve the equation for tan(3pi/8)

then we will go about using the half angle identity above to find tan(3pi/16)

Answer 15

okay

Answer 16

let me know if you want me to do the first round as an example

Answer 17

the second round is basically the same task

Answer 18

I got this so far:
\[\tan^2(\frac{ 3\pi }{ 8 })-2\tan(\frac{ 3\pi }{ 8 })-1=0\]

Answer 19

that is awesome

Answer 20

that is quadratic equation in terms of tan(3pi/8)

Answer 21

you can use the quadratic formula or complete the square to solve q

Answer 22

I'm not sure how to do that in terms of tan(3pi/8)

Answer 23

pretend v is tan(3pi/8)
\[v^2-2v-1=0 \\ v^2-2v=1 \\ v^2-2v+1=1+1 \\ (v-1)^2=2 \\ v-1= \pm \sqrt{2} \\ v=1 \pm \sqrt{2} \\ \text{ but remember } \tan(\frac{3\pi}{8}) \text{ is positive since } 0 < \frac{3\pi}{8} < \frac{\pi}{2} \\ \\ \text{ so } v=1+\sqrt{2} \\ \text{ or since } v=\tan(\frac{3\pi}{8}) \\ \tan(\frac{3\pi}{8})=1+\sqrt{2} \\ \\ \text{ now you need to look at solving this equation for } \tan(\frac{3\pi}{16}) \\ \tan(\frac{3\pi}{8})=\frac{2 \tan(\frac{3\pi}{16})}{1-\tan^2(\frac{3\pi}{16})}\]

Answer 24

Oh I see, so than it would be
\[1+\sqrt2=\frac{ 2\tan(3\pi/16) }{ 1-\tan^2(3\pi/16) }\]

Answer 25

this quadratic will turn out uglier but it is still doable

Answer 26

I need some help...

Answer 27

\[(1+\sqrt{2})(1)-(1+\sqrt{2})\tan^2(\frac{3\pi}{16})=2 \tan(\frac{3\pi}{16}) \\ -(1+\sqrt{2}) \tan^2(\frac{3\pi}{16})-2 \tan(\frac{3\pi}{16})+(1+\sqrt{2})=0 \\ \text{ \let } x=\tan(\frac{3\pi}{16}) \\ -(1+\sqrt{2})x^2-2 x+(1+\sqrt{2})=0 \]
a is that funny thing in front of x^2
b is the thing in front of x
the thingy left over, the constant part) is c

Answer 28

I would just plug into the quadratic formula from there

Answer 29

\[x=\frac{-(-2) \pm \sqrt{(-2)^2+4(1+\sqrt{2})(1+\sqrt{2})}}{-2(1+\sqrt{2})}\]

Answer 30

do you have something like this yet?

Answer 31

since again 3pi/16 is between 0 and pi/2
we must choose between the + or the minus in the numerator

we want our tan(3pi/16) to be positive
and since the bottom is negative we need our top to be negative
so we are going to go with the -

Answer 32

you can simplify that answer you have above too

Answer 33

Sorry for the wait. I got this:
\[\tan(3\pi/16)=\frac{ -1+\sqrt{4+2\sqrt{2}} }{ 1+\sqrt2 }\] ??

Answer 34

yep yep

Answer 35

and if you wanted to check it
that fraction on the right hand side gives:
http://www.wolframalpha.com/input/?i=%28-1%2Bsqrt%284%2B2+sqrt%282%29%29%29%2F%281%2Bsqrt%282%29%29


and tan(3pi/16) gives http://www.wolframalpha.com/input/?i=tan%283pi%2F16%29

Answer 36

both say .6681786 blah blah

Answer 37

Thanks! @freckles :)