Question Number 107672 by bemath last updated on 12/Aug/20
$$\:\:\:\:\:\spadesuit\mathcal{B}{e}\mathcal{M}{ath}\spadesuit \\ $$$$\:\:\:\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)}\:? \\ $$
Commented by Her_Majesty last updated on 12/Aug/20
$$\mathrm{2} \\ $$
Commented by ajfour last updated on 12/Aug/20
$$\:{Her}\:{MJS}\:{ty}\:\:{would}\:{you}\:{please}\:{explain} \\ $$$${this}\:{too}. \\ $$
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](https://www.tinkutara.com/question/Q107750.png)
$$\mathrm{We}\:\mathrm{prove}\:\mathrm{that}\:\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)}=\mathrm{2}\left(\ast\right) \\ $$$$\Leftrightarrow\mathrm{tan}\left(\frac{\pi}{\mathrm{7}}\right)\left[\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)\right]=\mathrm{2} \\ $$$$\Leftrightarrow\mathrm{4sin}\frac{\pi}{\mathrm{7}}\mathrm{cos}\frac{\pi}{\mathrm{14}}\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}+\mathrm{sin}\frac{\pi}{\mathrm{7}}=\mathrm{2cos}\frac{\pi}{\mathrm{7}}\mathrm{cos}\frac{\pi}{\mathrm{14}} \\ $$$$\Leftrightarrow\mathrm{2}\left(\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{14}}+\mathrm{sin}\frac{\pi}{\mathrm{14}}\right)\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}+\mathrm{sin}\frac{\pi}{\mathrm{7}} \\ $$$$=\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{14}}+\mathrm{cos}\frac{\pi}{\mathrm{14}} \\ $$$$\Leftrightarrow\left(\mathrm{2sin}\frac{\mathrm{3}\pi}{\mathrm{14}}\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\left(\mathrm{2sin}\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{sin}\frac{\pi}{\mathrm{14}}\right)\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}+\mathrm{sin}\frac{\pi}{\mathrm{7}}−\left(\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{14}}+\mathrm{cos}\frac{\pi}{\mathrm{14}}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{cos}\frac{\pi}{\mathrm{14}}−\mathrm{cos}\frac{\mathrm{7}\pi}{\mathrm{14}}+\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{14}}−\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{14}}+\mathrm{sin}\frac{\pi}{\mathrm{7}}+\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}−\left(\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{14}}+\mathrm{cos}\frac{\pi}{\mathrm{14}}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{sin}\frac{\pi}{\mathrm{7}}−\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{14}}=\mathrm{0}\left(\mathrm{since}\:\mathrm{cos}\frac{\mathrm{7}\pi}{\mathrm{14}}=\mathrm{cos}\frac{\pi}{\mathrm{2}}=\mathrm{0}\right) \\ $$$$\mathrm{This}\:\:\mathrm{equlity}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true}\:\mathrm{since} \\ $$$$\mathrm{sin}\frac{\pi}{\mathrm{7}}=\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{7}}\right)=\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{14}} \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{equality}\:\left(\ast\right)\mathrm{proved} \\ $$