$\tan (θ) = 1 \pm 2^{0.5}$

By Anonymous on Saturday, May 5, 2001 - 06:50 pm :

How would you find the values of

θ

\tan (θ) = 1 \pm 2^{0.5}

Thanks

By Brad Rodgers (P1930) on Sunday, May 6, 2001 - 03:14 am :

The answer to the 'plus one' is

3 π / 8

, and I haven't yet figured any sort of a solution to the second one in terms of pi. Here's a relatively easy way to prove that

\tan (3 π / 8) = 1 + 2^{1 / 2}

out darn tangents! out!
See if you can figure out why the angles are as they are. I'll for now assume you're okay with the way the angles are labeled. First, use the angle bisector theorem to deduce that

$1 / a = 2^{1 / 2} / b$

$b = 2^{1 / 2} a$

Remembering that

$a + b = 1$ , you should be able to substitute and see that $a = 1 / (1 + 2^{1 / 2})$ Thus the tangent of $3 π / 8 = 1 / [1 / (1 + 2^{1 / 2})] = 1 + 2^{1 / 2}$ as required. I'm still working on an angle that has tangent of $1 - 2^{1 / 2}$ , it seems to be $- π / 8$ , but I'm having trouble proving this.

Brad

By Brad Rodgers (P1930) on Sunday, May 6, 2001 - 03:27 am :

The proof is actually easier for the negative solution. The answer is

- π / 8

and you get there by noticing that

angle bisector
in the preceding triangle, we get, as before,

$b = 2^{1 / 2} a$

and

$a + b = 1$

So $a = 1 / (1 + 2^{1 / 2})$ or $2^{1 / 2} - 1$ . Therefore

$\tan (π / 8) = 2^{1 / 2} - 1$
Which, recalling that tan(-x)=-tan(x), gives us the answer. To get the full answer then, just remember that tan oscillates every pi.

Brad

By Kerwin Hui (Kwkh2) on Sunday, May 6, 2001 - 03:14 pm :

Another way of solving:

$\tan θ = 1 \pm \sqrt{2}$

so $(\tan θ - 1)^{2} = 2$

i.e. $\tan^{2} θ - 2 \tan θ - 1 = 0$

Now rearrange to get

$2 \tan θ / (1 - \tan^{2} θ) = - 1$

But we recognise the left hand side is $\tan 2 θ$ , and proceed from there.

Kerwin

By Anonymous on Sunday, May 6, 2001 - 08:22 pm :

Thanks Brad and Kerwin