prove-that-tan-3pi-7-4sin-pi-7-7-

Question Number 69081 by myintkhaing1121960@gmail.com last updated on 19/Sep/19

p r o v e t h a t

t a n \frac{3 π}{7} - 4 s i n \frac{π}{7} = \sqrt{7} .

Answered by Tanmay chaudhury last updated on 19/Sep/19

7 a = π

t a n 7 a = t a n π = - 0

\frac{s_{1} - s_{3} + s_{5} - s_{7}}{1 - s_{2} + s_{4} - s_{6}} = 0

7 c_{1} t a n a - 7 c_{3} {t a n}^{3} a + 7 c_{5} {t a n}^{5} a - 7 c_{7} {t a n}^{7} a = 0

7 t a n a - \frac{7 \times 6 \times 5}{3 \times 2} {t a n}^{3} a + \frac{7 \times 6}{2} {t a n}^{5} a - {t a n}^{7} a = 0

7 t a n a - 35 {t a n}^{3} a + 21 {t a n}^{5} a - {t a n}^{7} a = 0

t a n a = t a n \frac{π}{7} \neq 0

7 - 35 {t a n}^{2} a + 21 {t a n}^{4} a - {t a n}^{6} a = 0

{t a n}^{6} a - 21 {t a n}^{4} a + 35 {t a n}^{2} a - 7 = 0

f r o m a b o v e e q n w e h a v e t o f i n d v a l u e o f t a n a

t a n 3 a = \frac{3 t a n a - {t a n}^{3} a}{1 - 3 {t a n}^{2} a}

s i n a

= \sqrt{1 - \frac{1}{{s e c}^{2} a}}

= \sqrt{\frac{{t a n}^{2} a}{1 + {t a n}^{2} a}}

t a n 3 a - 4 s i n a

= \frac{3 t a n a - {t a n}^{3} a}{1 - 3 {t a n}^{2} a} - \frac{4 t a n a}{\sqrt{1 + {t a n}^{2} a}}

w a i t \dots

Answered by mind is power last updated on 19/Sep/19

w e h a v e

\underset{i = 1}{\prod^{m}} t a n (\frac{i π}{2 m + 1}) = \sqrt{2 m + 1}

w e t r y t o o a p p l i e t h i s g o r m = 3

\Rightarrow \tan (\frac{π}{7}) \tan (\frac{2 π}{7}) \tan (\frac{3 π}{7}) = \sqrt{7}

w e f i n i s h i f a n d o n l y i f

\Rightarrow \tan (\frac{π}{7}) \tan (\frac{2 π}{7}) \tan (\frac{3 π}{7}) = \tan (\frac{3 π}{7}) - 4 \sin (\frac{π}{7})

l e t a = \frac{π}{7}

w e w a n t t o o s h o w

\tan (a) \tan (2 a) \tan (3 a) = \tan (3 a) - 4 \sin (a)

\Leftrightarrow \frac{\sin (a) \sin (2 a) \sin (3 a)}{\cos (a) \cos (2 a) \cos (3 a)} = \frac{\sin (3 a) - 4 \sin (a) \cos (3 a)}{\cos (3 a)}

\Leftrightarrow \frac{\sin (a) . 2 \sin (a) \cos (a) \sin (3 a)}{\cos (a) \cos (2 a) \cos (3 a)} = \frac{\sin (3 a) - 2 . 2 \sin (a) c o s (3 a)}{\cos (3 a)}

\Leftrightarrow \frac{2 \sin^{2} (a) \sin (3 a)}{\cos (2 a)} = \sin (3 a) - 2 . 2 \cos (3 a) \sin (a)

\sin (x) \cos (y) = \frac{s i n (x + y) + s i n (x - y)}{2} \Rightarrow 2 \sin (a) \cos (3 a) = s i n (4 a) - s i n (2 a)

2 {s i n}^{2} (a) = 1 - c o s (2 a)

\Rightarrow

\frac{2 \sin^{2} (a) \sin (3 a)}{\cos (2 a)} = \sin (3 a) - 2 . 2 \cos (3 a) \sin (a)

\Leftrightarrow \frac{s i n (3 a) (1 - c o s (2 a))}{c o s (2 a)} = s i n (3 a) - 2 s i n (4 a) + 2 s i n (2 a)

\Leftrightarrow s i n (3 a) - s i n (3 a) c o s (2 a) = s i n (3 a) c o s (2 a) - 2 s i n (4 a) c o s (2 a) + 2 s i n (2 a) c o s (2 a)

\Leftrightarrow s i n (3 a) - \frac{s i n (5 a) + s i n (a)}{2} = \frac{s i n (5 a) + s i n (a)}{2} - s i n (6 a) - s i n (2 a) + s i n (4 a)

s i n (4 a) = s i n (3 a) \dots . a = \frac{π}{7}

s i n (6 a) = s i n (a)

s i n (5 a) = s i n (2 a)

\Rightarrow s i n (3 a) - \frac{s i n (2 a) + s i n (a)}{2} = \frac{s i n (2 a) + s i n (a)}{2} - s i n (a) - s i n (2 a) + s i n (3 a)

\Rightarrow 0 = \frac{s i n (2 a) + s i n (a) + s i n (2 a) + s i n (a)}{2} - s i n (a) - s i n (2 a) - s i n (3 a) + s i n (3 a)

\Leftrightarrow 0 = 0

\Rightarrow t a n (\frac{3 π}{7}) - 4 s i n (\frac{π}{7}) = t a n (\frac{π}{7}) t a n (\frac{2 π}{7}) t a n (\frac{3 π}{7}) = \underset{i = 1}{\prod^{3}} \tan (\frac{i π}{7}) = \sqrt{7}

Commented by MJS last updated on 19/Sep/19

great!

Commented by mind is power last updated on 19/Sep/19

t h a n k y o u