Question

Solve for value of tan3pi/16
\[1+\sqrt2=\frac{ 2\tan \frac{ 3\pi }{ 16 } }{ 1-\tan^2\frac{ 3\pi }{ 16 } }\]

Answer 1

So basically you just have \(1+\sqrt2=\dfrac{2x}{1-x^2}\)

Answer 2

So I move the denominator to the left side?

Answer 3

Yeah I would do that.

\((1+\sqrt2)(x^2-1) = 2x\)

Answer 4

Arrange a little more, then you can apply quadratic formula, I guess.

Answer 5

Do I expand it first?

Answer 6

How can I get the answer to exact values?

Answer 7

Nah just distribute \((1+\sqrt2)\)

You want equation to be in form \(ax^2+bx+c=0\)

So here \(-(1+\sqrt2)x^2-2x+(1+\sqrt2)\)

So now we can apply quadratic formula:

\(a=-(1+\sqrt2)\\b=-2\\c=1+\sqrt2\)

Answer 8

\[x = \dfrac{-b+\sqrt{b^2-4ac}}{2a}\]

Answer 9

Hmm this process seems pretty beastly :)
Do you remember the Tangent Double Angle Formula we used in the last problem Sammy? You probably should start with that. Just another idea.

Answer 10

I'm not getting this.. :(

Answer 11

Hmm, yeah this method is much more preferable.

Tangent Double Angle Formula: \(\displaystyle \tan(2x) = \dfrac{2\tan(x)}{1-\tan^2(x)}\)

Answer 12

What do I do with the tan(2x) identity?

Answer 13

Oh I got it now! @zepdrix :) I kinda used the quadratic formula. I took me a long time to figure it out!

Answer 14

Oh you figure it out? :D
Sorry I thought pony was helping you.