Question 150d7

Jun 12, 2015

Show cos(2arctan (1/7)) = sin (4arctan (1/3)#

Explanation:

$\tan x = \frac{1}{7} - \to x = \arctan \left(\frac{1}{7}\right) = 8.13 \to 2 x = 16.26$

$\cos 2 x = \cos 16.26 = 0.96$

$\tan y = \frac{1}{3} - \to y = \tan 18.43 \to 4 y = 73.74$ deg

$\sin 4 y = \sin 73.74 = 0.96$

Jun 12, 2015

See the proof:

Explanation:

We have to prove that

$\cos \alpha = \sin \beta$, where

$\alpha = 2 \arctan \left(\frac{1}{7}\right)$ and

$\beta = 4 \arctan \left(\frac{1}{3}\right)$.

So:

$\tan \left(\frac{\alpha}{2}\right) = \frac{1}{7}$

and

$\tan \left(\frac{\beta}{4}\right) = \frac{1}{3}$.

Remembering the double-angle formulae:

$\sin x = \frac{2 t}{1 + {t}^{2}}$,

$\cos x = \frac{1 - {t}^{2}}{1 + {t}^{2}}$ where $t = \tan \left(\frac{x}{2}\right)$

and

$\sin 2 x = 2 \sin x \cos x$,

then:

$\cos \alpha = 2 \sin \left(\frac{\beta}{2}\right) \cos \left(\frac{\beta}{2}\right) \Rightarrow$

(where ${t}_{1} = \tan \left(\frac{\alpha}{2}\right)$ and ${t}_{2} = \tan \left(\frac{\beta}{4}\right)$

$\frac{1 - {t}_{1}^{2}}{1 + {t}_{1}^{2}} = 2 \cdot \frac{2 {t}_{2}}{1 + {t}_{2}^{2}} \cdot \frac{1 - {t}_{2}^{2}}{1 + {t}_{2}^{2}}$

$\frac{1 - \frac{1}{49}}{1 + \frac{1}{49}} = 2 \cdot \frac{2 \cdot \frac{1}{3}}{1 + \frac{1}{9}} \cdot \frac{1 - \frac{1}{9}}{1 + \frac{1}{9}} \Rightarrow$

$\frac{\frac{48}{49}}{\frac{50}{49}} = 2 \cdot \frac{\frac{2}{3}}{\frac{9}{10}} \cdot \frac{\frac{8}{9}}{\frac{9}{10}} \Rightarrow$

$\frac{48}{49} \cdot \frac{49}{50} = 2 \cdot \frac{2}{3} \cdot \frac{9}{10} \cdot \frac{8}{9} \cdot \frac{9}{10} \Rightarrow$

$\frac{24}{25} = \frac{24}{25}$.