Find-tan-142-5-without-tables-

Question Number 173447 by ajfour last updated on 11/Jul/22

F i n d \tan (142.5 °) w i t h o u t

t a b l e s .

Commented by mr W last updated on 12/Jul/22

w e l c o m e b a c k s i r!

l e t t = \tan 15 °

\tan 30 ° = \frac{2 t}{1 - t^{2}} = \frac{1}{\sqrt{3}}

t^{2} + 2 \sqrt{3} t - 1 = 0

\Rightarrow t = 2 - \sqrt{3}

l e t s = \tan 142.5 °

\frac{1}{\tan 15 °} = \tan 75 ° = - \tan 285 ° = - \tan (2 \times 142.5 °)

\frac{1}{2 - \sqrt{3}} = - \frac{2 s}{1 - s^{2}}

s^{2} - 2 (2 - \sqrt{3}) s - 1 = 0

\Rightarrow s = \tan 142.5 ° = 2 - \sqrt{3} - 2 \sqrt{2 - \sqrt{3}}

= (\sqrt{2} - \sqrt{3}) (\sqrt{2} + 1)

Commented by ajfour last updated on 11/Jul/22

\sqrt{2 - \sqrt{3}} = p

2 - \sqrt{3} = p^{2} = \frac{1}{2 + \sqrt{3}}

p^{2} + \frac{1}{p^{2}} = 4

{(p + \frac{1}{p})}^{2} = 6

p + \frac{1}{p} = \sqrt{6}

p^{2} - \sqrt{6} p + 1 = 0

p = \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}

h e n c e r e w r i t i n g a n o t h e r w a y

s = 2 - \sqrt{3} - 2 p

= 2 - \sqrt{3} - \sqrt{6} + \sqrt{2}

t h a n k y o u s i r .

Commented by Tawa11 last updated on 11/Jul/22

Great sirs

Commented by Tawa11 last updated on 13/Jul/22

Great sir

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