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

     ♠BeMath♠      ((4sin (((2π)/7))+sec ((π/(14))))/(cot ((π/7)))) ?

$$\:\:\:\:\:\spadesuit\mathcal{B}{e}\mathcal{M}{ath}\spadesuit \\ $$$$\:\:\:\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)}\:? \\ $$

Commented by Her_Majesty last updated on 12/Aug/20

2

$$\mathrm{2} \\ $$

Commented by ajfour last updated on 12/Aug/20

 Her MJS ty  would you please explain  this too.

$$\:{Her}\:{MJS}\:{ty}\:\:{would}\:{you}\:{please}\:{explain} \\ $$$${this}\:{too}. \\ $$

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

We prove that  ((4sin (((2π)/7))+sec ((π/(14))))/(cot ((π/7))))=2(∗)  ⇔tan((π/7))[4sin (((2π)/7))+sec ((π/(14)))]=2  ⇔4sin(π/7)cos(π/(14))sin((2π)/7)+sin(π/7)=2cos(π/7)cos(π/(14))  ⇔2(sin((3π)/(14))+sin(π/(14)))sin((2π)/7)+sin(π/7)  =cos((3π)/(14))+cos(π/(14))  ⇔(2sin((3π)/(14))sin((2π)/7))+(2sin((2π)/7)sin(π/(14)))sin((2π)/7)+sin(π/7)−(cos((3π)/(14))+cos(π/(14)))=0  ⇔cos(π/(14))−cos((7π)/(14))+cos((3π)/(14))−cos((5π)/(14))+sin(π/7)+sin((2π)/7)−(cos((3π)/(14))+cos(π/(14)))=0  ⇔sin(π/7)−cos((5π)/(14))=0(since cos((7π)/(14))=cos(π/2)=0)  This  equlity is always true since  sin(π/7)=cos((π/2)−(π/7))=cos((5π)/(14))  Thus,the equality (∗)proved

$$\mathrm{We}\:\mathrm{prove}\:\mathrm{that}\:\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)}=\mathrm{2}\left(\ast\right) \\ $$$$\Leftrightarrow\mathrm{tan}\left(\frac{\pi}{\mathrm{7}}\right)\left[\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)\right]=\mathrm{2} \\ $$$$\Leftrightarrow\mathrm{4sin}\frac{\pi}{\mathrm{7}}\mathrm{cos}\frac{\pi}{\mathrm{14}}\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}+\mathrm{sin}\frac{\pi}{\mathrm{7}}=\mathrm{2cos}\frac{\pi}{\mathrm{7}}\mathrm{cos}\frac{\pi}{\mathrm{14}} \\ $$$$\Leftrightarrow\mathrm{2}\left(\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{14}}+\mathrm{sin}\frac{\pi}{\mathrm{14}}\right)\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}+\mathrm{sin}\frac{\pi}{\mathrm{7}} \\ $$$$=\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{14}}+\mathrm{cos}\frac{\pi}{\mathrm{14}} \\ $$$$\Leftrightarrow\left(\mathrm{2sin}\frac{\mathrm{3}\pi}{\mathrm{14}}\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\left(\mathrm{2sin}\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{sin}\frac{\pi}{\mathrm{14}}\right)\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}+\mathrm{sin}\frac{\pi}{\mathrm{7}}−\left(\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{14}}+\mathrm{cos}\frac{\pi}{\mathrm{14}}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{cos}\frac{\pi}{\mathrm{14}}−\mathrm{cos}\frac{\mathrm{7}\pi}{\mathrm{14}}+\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{14}}−\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{14}}+\mathrm{sin}\frac{\pi}{\mathrm{7}}+\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}−\left(\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{14}}+\mathrm{cos}\frac{\pi}{\mathrm{14}}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{sin}\frac{\pi}{\mathrm{7}}−\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{14}}=\mathrm{0}\left(\mathrm{since}\:\mathrm{cos}\frac{\mathrm{7}\pi}{\mathrm{14}}=\mathrm{cos}\frac{\pi}{\mathrm{2}}=\mathrm{0}\right) \\ $$$$\mathrm{This}\:\:\mathrm{equlity}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true}\:\mathrm{since} \\ $$$$\mathrm{sin}\frac{\pi}{\mathrm{7}}=\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{7}}\right)=\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{14}} \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{equality}\:\left(\ast\right)\mathrm{proved} \\ $$

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