Exact value of arctan[1+sqrt(2)]

In fact all you need is the first identity.

The second identity is a hint to the easier way to express it as $\tan\frac{\pi}{2^n} = \sqrt{\frac{2 - \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}}{2 + \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}}}$ where there are $n - 1$ twos under the radical.

The second identity is a hint to the easier way to express it as $\tan\frac{\pi}{2^n} = \sqrt{\frac{2 - \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}}{2 + \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}}}$ where there are $n - 1$ twos under the radical.

In fact all you need is the first identity. If you rearrange

$\tan 2\theta = \dfrac{2\tan \theta}{1-\tan^2 \theta}$

then, using the quadratic formula, you get

$\tan \theta = \dfrac{-1 \pm \sqrt{1+\tan^2 2\theta}}{\tan 2\theta}$

If we write $t_n = \tan \dfrac{\pi}{2^n}$ (for $n \ge 2$), then we must have $t_n > 0$ and so we take the + sign, and the above equation gives

$t_n = \dfrac{-1 + \sqrt{1+t_{n-1}^2}}{t_{n-1}}$

which is quite nice.

Sorry if am being very silly.

I don't really get what you mean by attaching this image?

What is its significance?

I don't really get what you mean by attaching this image?

What is its significance?

A person of your ability does not need an explanation.

Think about it.

Consider the bottom side as value 1. Work out the rest of the lengths and angles and you'll see what he meant.

I do understand till the rearrangement and applying the quadratic formula, but after it i am not really getting how are you doing it. Where does this $t_{n-1}$ comes from?

Sorry if am being very silly.

If $t_n = \tan \dfrac{\pi}{2^n}$ then set $\theta = \dfrac{\pi}{2^n}$ in the identity. You get $2\theta = \dfrac{\pi}{2^{n-1}}$ and so $\tan 2\theta = t_{n-1}$.

A few questions about this,

Shouldn't the range be $n > 2$ rather than $n \ge 2$
Inputting $n=2$ gives,
$t_2 = \dfrac{-1+\sqrt{1+t_{1}^{2}}}{t_1}$
$t_1 = \tan \dfrac{\pi}{2} = \text{math error}$

The other question is, i don't completely understand why you rejected the negative sign.

you're not a calculator!

That's what I meant. Though I'd try to avoid saying things like "= math error"... you're not a calculator!


The graph of $y=\tan x$ is nonnegative whenever $0 \le \theta < \dfrac{\pi}{2}$.

you're not a calculator!

An easy way to find tan(3pi/8):

Consider the triangle ABC where A is the origin, B is what you get by starting at A and going 2 unit with bearing 0, and C is what you get by starting from B and going 2 unit with bearing pi/4.

In other words, A = (0,0), B = (0, 2) and C=(sqrt(2), 2+sqrt(2)). Then AC has bearing 3pi/4 and gradient 1+sqrt(2).

I remember in the AEA it was a common thing to show tan(pi/8) = sqrt(2) - 1 (then the answer for this comes quite quickly from that), but I don't quite remember how to do it