BeMath-4sin-2pi-7-sec-pi-14-cot-pi-7-

Question Number 107672 by bemath last updated on 12/Aug/20

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\frac{4 \sin (\frac{2 π}{7}) + \sec (\frac{π}{14})}{\cot (\frac{π}{7})} ?

Commented by Her_Majesty last updated on 12/Aug/20

2

Commented by ajfour last updated on 12/Aug/20

H e r M J S t y w o u l d y o u p l e a s e e x p l a i n

t h i s t o o .

Answered by 1549442205PVT last updated on 12/Aug/20

We prove that \frac{4 \sin (\frac{2 π}{7}) + \sec (\frac{π}{14})}{\cot (\frac{π}{7})} = 2 (*)

\Leftrightarrow \tan (\frac{π}{7}) [4 \sin (\frac{2 π}{7}) + \sec (\frac{π}{14})] = 2

\Leftrightarrow 4 \sin \frac{π}{7} \cos \frac{π}{14} \sin \frac{2 π}{7} + \sin \frac{π}{7} = 2 \cos \frac{π}{7} \cos \frac{π}{14}

\Leftrightarrow 2 (\sin \frac{3 π}{14} + \sin \frac{π}{14}) \sin \frac{2 π}{7} + \sin \frac{π}{7}

= \cos \frac{3 π}{14} + \cos \frac{π}{14}

\Leftrightarrow (2 \sin \frac{3 π}{14} \sin \frac{2 π}{7}) + (2 \sin \frac{2 π}{7} \sin \frac{π}{14}) \sin \frac{2 π}{7} + \sin \frac{π}{7} - (\cos \frac{3 π}{14} + \cos \frac{π}{14}) = 0

\Leftrightarrow \cos \frac{π}{14} - \cos \frac{7 π}{14} + \cos \frac{3 π}{14} - \cos \frac{5 π}{14} + \sin \frac{π}{7} + \sin \frac{2 π}{7} - (\cos \frac{3 π}{14} + \cos \frac{π}{14}) = 0

\Leftrightarrow \sin \frac{π}{7} - \cos \frac{5 π}{14} = 0 (since \cos \frac{7 π}{14} = \cos \frac{π}{2} = 0)

This equlity is always true since

\sin \frac{π}{7} = \cos (\frac{π}{2} - \frac{π}{7}) = \cos \frac{5 π}{14}

Thus, the equality (*) proved