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Question Number 80595 by Power last updated on 04/Feb/20

Commented by Power last updated on 04/Feb/20

$prove that$
Commented by Power last updated on 04/Feb/20

$sir W pls$
	Answered by mind is power last updated on 05/Feb/20

	$m y i d e e s t a r t$ $w i t h e$ $t g (3.10) = t g (30), t g (3.50) = t g (150) = t g (180 - 30) = - t g (30)$ $t g (3.70) = t g (210) = t g (180 + 30) = t g (30) = \frac{1}{\sqrt{3}}$ $\Rightarrow - t g (50), t g (10), t g (70) R o o t o f S o m m p o l y n o m i a l i n t a n (a)$ $t a n (3 a) = \frac{t a n (a) + t a n (2 a)}{1 - t a n (a) t a n (2 a)}$ $t a n (2 a) = \frac{2 t a n (a)}{1 - {t a n}^{2} (a)} \Rightarrow t a n (3 a) = \frac{3 t a n (a) - {t a n}^{3} (a)}{1 - 3 {t a n}^{2} (a)}$ $(- t a n (50), t a n (10), t a n (70)) R o o t o f$ $\frac{3 X - X^{3}}{1 - 3 X^{2}} = \frac{1}{\sqrt{3}} \Leftrightarrow X^{3} - \sqrt{3} X - 3 X + \frac{1}{\sqrt{3}} = 0$ $a = t a n (10), b = t a n (70), c = - t a n (50)$ $V i e t i d e n t i t i e s \Rightarrow$ $a + b + c = \sqrt{3} = e_{1}$ $e_{2} = a b + b c + c a = - 3$ $e_{3} = a b c = - \frac{1}{\sqrt{3}}, e_{n} = 0, \forall n ⩾ 4$ $w e u s e n e w t o o n i d e n t i t i e s$ $l e t p_{k} = a^{k} + b^{k} + c^{k}$ $p_{1} = e_{1} = \sqrt{3}$ $p_{2} = e_{1} p_{1} - 2 e_{2} = 9$ $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3} = 11 \sqrt{3}$ $p_{4} = e_{1} p_{3} - e_{2} p_{2} + e_{3} p_{1} = 59$ $p_{5} = e_{1} p_{4} - e_{2} p_{3} + e_{3} p_{2} = 89 \sqrt{5}$ $p_{6} = e_{1} p_{5} - e_{2} p_{4} + e_{3} p_{3} = 89 \sqrt{3} . \sqrt{3} + 3.59 - \frac{1}{\sqrt{3}} . 11 \sqrt{3} = 433$ $p_{6} = a^{6} + b^{6} + c^{6} = {t g}^{6} (10) + {(- t g (50))}^{6} + {t g}^{6} (70) = 433$ $= {t g}^{6} (10) + {t g}^{6} (50) + {t g}^{6} (70) = 433 W e g e t o u r r e s u l t s$
	Commented by Power last updated on 05/Feb/20

	$thank you sir . Great!$
	Commented by mind is power last updated on 05/Feb/20

	$w i t h e p l e a s u r$
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