Math Question and Answers


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 210677 by Safojon last updated on 16/Aug/24

	Answered by BHOOPENDRA last updated on 16/Aug/24

	$θ = \frac{k π}{7}, 7 θ = k π (w h e r e k = 1, 2, 3, 4 . . . . n)$ $7 θ = π (k = 1)$ $4 θ + 3 θ = π$ $t g 4 θ = - t g (3 θ) l e t t g = t$ $\Rightarrow \frac{4 t - 4 t^{3}}{1 - 6 t^{2} + t^{4}} = \frac{- 3 t - t^{3}}{1 - 3 t^{2}}$ $= t^{6} - 21 t^{4} + 35 t^{2} - 7 = 0 e q (1)$ $= T h i s i s a c u b i c e q u a t i o n i n t^{2} t h a t$ $i s i n {t g}^{2} θ .$ $t h e r o o t o f t h e e q i s {t g}^{2} \frac{π}{7}, {t g}^{2} \frac{2 π}{7}$ ${t g}^{2} \frac{3 π}{7}$ $S o s u m o f t h e r o o t = - \frac{(- 21)}{1} = 21$ ${t g}^{2} \frac{π}{7} + {t g}^{2} \frac{2 π}{7} + {t g}^{2} \frac{3 π}{7} = 21$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum