(i)((1/2)−cos (π/7))((1/2)−cos ((3π)/7))((1/2)−cos ((9π)/7))? (ii) ((√3)+tan 1°)((√3)+tan 2°)((√3)+tan 3°)×...×((√3)+tan 29°)?


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Question Number 119937 by bobhans last updated on 28/Oct/20

$(i) (\frac{1}{2} - \cos \frac{π}{7}) (\frac{1}{2} - \cos \frac{3 π}{7}) (\frac{1}{2} - \cos \frac{9 π}{7}) ?$ $(i i) (\sqrt{3} + \tan 1 °) (\sqrt{3} + \tan 2 °) (\sqrt{3} + \tan 3 °) \times . . . \times (\sqrt{3} + \tan 29 °) ?$
	Answered by TANMAY PANACEA last updated on 28/Oct/20

	$i i) \sqrt{3} + t a n 1^{o}$ $= t a n 60^{o} + t a n 1^{o}$ $= \frac{s i n 60 c o s 1 + c o s 6 o s i n 1}{c o s 60 c o s 1} = \frac{s i n (61)}{c o s 60 c o s 1}$ $n o w$ $p = \frac{s i n 61 \times s i n 62 \times s i n 63 \times . . . \times s i n 89}{{(c o s 60)}^{29} c o s 1 c o s 2 c o s 3 . . . c o s 29}$ $= 2^{29}$ $=$
	Answered by TANMAY PANACEA last updated on 28/Oct/20

	$7 θ = π$ $c o s 3 θ = c o s (π - 4 θ)$ $4 {c o s}^{3} θ - 3 c o s θ = - (2 {c o s}^{2} 2 θ - 1)$ $4 {c o s}^{3} θ - 3 c o s θ = - 2 {(2 {c o s}^{2} θ - 1)}^{2} + 1$ $4 c^{3} - 3 c = - 2 (4 c^{4} - 4 c^{2} + 1) + 1$ $4 c^{3} - 3 c = - 8 c^{4} + 8 c^{2} - 2 + 1$ $8 c^{4} + 4 c^{3} - 8 c^{2} - 3 c + 1 = 0$ $⇛ 8 c^{4} + 8 c^{3} - 4 c^{3} - 4 c^{2} - 4 c^{2} - 4 c + c + 1 = 0$ $⇛ 8 c^{3} (c + 1) - 4 c^{2} (c + 1) - 4 c (c + 1) + 1 (c + 1) = 0$ $⇛ (c + 1) (8 c^{3} - 4 c^{2} - 4 c + 1) = 0$ $g i v e n p r o b l e m$ $(a - c_{1}) (a - c_{2}) (a - c_{3})$ $a^{3} - a^{2} (c_{1} + c_{2} + c_{3}) + a (c_{1} c_{2} + c_{2} c_{3} + c_{1} c_{3}) - c_{1} c_{2} c_{3}$ $a = \frac{1}{2} c_{1} = c o s \frac{π}{7} c_{2} = c o s \frac{3 π}{7} c_{3} = c o s \frac{9 π}{7}$ $c o s \frac{π}{7}, c o s \frac{3 π}{7} a n d c o s \frac{9 π}{7} s a t i s f y e q n$ $8 c^{3} - 4 c^{2} - 4 c + 1 = 0$ $S o a n s w e r i s$ $c_{1} + c_{2} + c_{3} = \frac{- (- 4)}{8}$ $c_{1} c_{2} + c_{2} c_{3} + c_{1} c_{3} = \frac{- 4}{8}$ $c_{1} c_{2} c_{3} = \frac{- 1}{8}$ $s o a n s w e r i s$ $a^{3} - a^{2} (c_{1} + c_{2} + c_{3}) + a (c_{1} c_{2} + c_{2} c_{3} + c_{1} c_{3}) - c_{1} c_{2} c_{3}$ $= \frac{1}{8} - \frac{1}{4} (\frac{1}{2}) + \frac{1}{2} (\frac{- 1}{2}) - (\frac{- 1}{8})$ $= \frac{1}{8} - \frac{1}{8} - \frac{1}{4} + \frac{1}{8}$ $= \frac{1 - 2}{8} = \frac{- 1}{8}$ $p l s c h k$
	Commented by Dwaipayan Shikari last updated on 28/Oct/20

	$Y e s i t i s c o r r e c t s i r!$
	Commented by TANMAY PANACEA last updated on 28/Oct/20

	$t h a n k y o u$
	Answered by bemath last updated on 28/Oct/20

	$(i i) (\sqrt{3} + \tan 1 °) (\sqrt{3} + \tan 29 °) = 3 + \sqrt{3} (\tan 29 ° + \tan 1 °) + \tan 29 ° . \tan 1 °$ $= 3 + \sqrt{3} (\tan 30 °) (1 - \tan 29 ° \tan 1 °) + \tan 29 ° . \tan 1 °$ $= 3 + 1 = 4$ $s i m i l a r l y (\sqrt{3} + \tan 2 °) (\sqrt{3} + \tan 28 °) = 4$ $t h e r e f o r e w e h a v e$ $= (4) (4) \times . . . (4) \times (\sqrt{3} + \tan 15 °)$ $= 2^{28} \times (\sqrt{3} + \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}) = 2^{28} \times (\sqrt{3} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1})$ $= 2^{28} \times (\sqrt{3} + \frac{4 - 2 \sqrt{3}}{2}) = 2^{28} \times (\sqrt{3} + 2 - \sqrt{3})$ $= 2^{29}$
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