i-1-2-cos-pi-7-1-2-cos-3pi-7-1-2-cos-9pi-7-ii-3-tan-1-3-tan-2-3-tan-3-3-tan-29-

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 \dots \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 \dots \times s i n 89}{{(c o s 60)}^{29} c o s 1 c o s 2 c o s 3 \dots 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 \dots (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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