Question

(tgπ/8) / (1-tg²π/8)

Answer 1

Отметьте пожалуйста лучшим решением и поставьте сердечко

Answer 2

To simplify the given expression, we'll use trigonometric identities and properties. Let's start by recalling some basic trigonometric identities:

1. $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$

Now, let's simplify the expression $\frac{\tan(\frac{\pi}{8})}{1 - \tan^2(\frac{\pi}{8})}$ using the above identity:

First, we'll find $\tan(\frac{\pi}{8})$:

$\tan(\frac{\pi}{8}) = \frac{\sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})}$

We can use the half-angle formulas to find $\sin(\frac{\pi}{8})$ and $\cos(\frac{\pi}{8})$:

$\sin(\frac{\pi}{8}) = \sqrt{\frac{1 - \cos(\frac{\pi}{4})}{2}} = \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{2}}$

$\cos(\frac{\pi}{8}) = \sqrt{\frac{1 + \cos(\frac{\pi}{4})}{2}} = \sqrt{\frac{1 + \frac{1}{\sqrt{2}}}{2}}$

Now, we can plug these values into $\tan(\frac{\pi}{8})$:

$\tan(\frac{\pi}{8}) = \frac{\sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})} = \frac{\sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{2}}}{\sqrt{\frac{1 + \frac{1}{\sqrt{2}}}{2}}}$

Next, we'll find $\tan^2(\frac{\pi}{8})$:

$\tan^2(\frac{\pi}{8}) = \left(\frac{\sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})}\right)^2 = \frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}$