Question #acd9c

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Question #acd9c

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Answer 1 · 2017-08-16T05:27:00

LHS of given identity

$= sec θ + sec 2 θ + sec 4 θ$ $=sectheta+sec2theta+sec4theta$

$= sec (\frac{2 π}{7}) + sec (\frac{4 π}{7}) + sec (\frac{8 π}{7})$ $=sec((2pi)/7)+sec((4pi)/7)+sec((8pi)/7)$

$= \frac{1}{cos (\frac{2 π}{7})} + \frac{1}{cos (\frac{4 π}{7})} + \frac{1}{cos (π + \frac{π}{7})}$ $=1/cos((2pi)/7)+1/cos((4pi)/7)+1/cos(pi+pi/7)$

$= \frac{1}{cos (\frac{2 π}{7})} + \frac{1}{cos (\frac{4 π}{7})} - \frac{1}{cos (\frac{π}{7})}$ $=1/cos((2pi)/7)+1/cos((4pi)/7)-1/cos(pi/7)$

$= \frac{cos (\frac{4 π}{7}) cos (\frac{π}{7}) + cos (\frac{2 π}{7}) cos (\frac{π}{7}) - cos (\frac{4 π}{7}) cos (\frac{2 π}{7})}{cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) cos (\frac{π}{7})}$ $=(cos((4pi)/7)cos(pi/7)+cos((2pi)/7)cos(pi/7)-cos((4pi)/7)cos((2pi)/7))/(cos((2pi)/7)cos((4pi)/7)cos(pi/7))$

$= \frac{2 cos (\frac{4 π}{7}) cos (\frac{π}{7}) + 2 cos (\frac{2 π}{7}) cos (\frac{π}{7}) - 2 cos (\frac{4 π}{7}) cos (\frac{2 π}{7})}{2 cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) cos (\frac{π}{7})}$ $=(2cos((4pi)/7)cos(pi/7)+2cos((2pi)/7)cos(pi/7)-2cos((4pi)/7)cos((2pi)/7))/(2cos((2pi)/7)cos((4pi)/7)cos(pi/7))$

$= \frac{cos (\frac{5 π}{7}) + cos (\frac{3 π}{7}) + cos (\frac{3 π}{7}) + cos (\frac{π}{7}) - cos (\frac{6 π}{7}) - cos (\frac{2 π}{7})}{2 cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) cos (\frac{π}{7})}$ $=(cos((5pi)/7)+cos((3pi)/7)+cos((3pi)/7)+cos(pi/7)-cos((6pi)/7)-cos((2pi)/7))/(2cos((2pi)/7)cos((4pi)/7)cos(pi/7))$

$= \frac{cos (π - \frac{2 π}{7}) + cos (\frac{3 π}{7}) + cos (\frac{3 π}{7}) + cos (\frac{π}{7}) - cos (π - \frac{π}{7}) - cos (\frac{2 π}{7})}{2 cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) cos (\frac{π}{7})}$ $=(cos(pi-(2pi)/7)+cos((3pi)/7)+cos((3pi)/7)+cos(pi/7)-cos(pi-pi/7)-cos((2pi)/7))/(2cos((2pi)/7)cos((4pi)/7)cos(pi/7))$

$= \frac{2 cos (\frac{3 π}{7}) + 2 cos (\frac{π}{7}) - 2 cos (\frac{2 π}{7})}{2 cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) cos (\frac{π}{7})}$ $=(2cos((3pi)/7)+2cos(pi/7)-2cos((2pi)/7))/(2cos((2pi)/7)cos((4pi)/7)cos(pi/7))$

$= \frac{cos (\frac{3 π}{7}) + cos (\frac{π}{7}) - cos (\frac{2 π}{7})}{cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) cos (\frac{π}{7})}$ $=(cos((3pi)/7)+cos(pi/7)-cos((2pi)/7))/(cos((2pi)/7)cos((4pi)/7)cos(pi/7))$

$= - \frac{cos (\frac{2 π}{7}) - cos (\frac{π}{7}) - cos (\frac{3 π}{7})}{cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) cos (\frac{π}{7})}$ $=-(cos((2pi)/7)-cos(pi/7)-cos((3pi)/7))/(cos((2pi)/7)cos((4pi)/7)cos(pi/7))$

Now the numerator

$cos (\frac{2 π}{7}) - cos (\frac{3 π}{7}) - cos (\frac{π}{7})$ $cos((2pi)/7)-cos((3pi)/7)-cos(pi/7)$

$= \frac{1}{2 sin (\frac{π}{7})} (2 sin (\frac{π}{7}) cos (\frac{2 π}{7}) - 2 sin (\frac{π}{7}) cos (\frac{3 π}{7}) - 2 sin (\frac{π}{7}) cos (\frac{π}{7}))$ $=1/(2sin(pi/7))(2sin(pi/7)cos((2pi)/7)-2sin(pi/7)cos((3pi)/7)-2sin(pi/7)cos(pi/7))$

$= \frac{1}{2 sin (\frac{π}{7})} (sin (\frac{3 π}{7}) - sin (\frac{π}{7}) - sin (\frac{4 π}{7}) + sin (\frac{2 π}{7}) - sin (\frac{2 π}{7}))$ $=1/(2sin(pi/7))(sin((3pi)/7)-sin(pi/7)-sin((4pi)/7)+sin((2pi)/7)-sin((2pi)/7))$

$= \frac{1}{2 sin (\frac{π}{7})} (sin (\frac{3 π}{7}) - sin (\frac{π}{7}) - sin (π - \frac{3 π}{7}))$ $=1/(2sin(pi/7))(sin((3pi)/7)-sin(pi/7)-sin(pi-(3pi)/7))$

$= \frac{1}{2 sin (\frac{π}{7})} (sin (\frac{3 π}{7}) - sin (\frac{π}{7}) - sin (\frac{3 π}{7}))$ $=1/(2sin(pi/7))(sin((3pi)/7)-sin(pi/7)-sin((3pi)/7))$

$= \frac{1}{2 sin (\frac{π}{7})} \times (- sin (\frac{π}{7}))$ $=1/(2sin(pi/7))xx(-sin(pi/7))$

$= - \frac{1}{2}$ $=-1/2$

And the denominator

$= (cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) cos (\frac{π}{7}))$ $=(cos((2pi)/7)cos((4pi)/7)cos(pi/7))$

$= \frac{8 sin (\frac{π}{7}) cos (\frac{2 π}{7}) cos (\frac{4 π}{7}) cos (\frac{π}{7})}{8 sin (\frac{π}{7})}$ $=(8sin(pi/7)cos((2pi)/7)cos((4pi)/7)cos(pi/7))/(8sin(pi/7))$

$= \frac{4 sin (\frac{2 π}{7}) cos (\frac{2 π}{7}) cos (\frac{4 π}{7})}{8 sin (\frac{π}{7})}$ $=(4sin((2pi)/7)cos((2pi)/7)cos((4pi)/7))/(8sin(pi/7))$

$= \frac{2 sin (\frac{4 π}{7}) cos (\frac{4 π}{7})}{8 sin (\frac{π}{7})}$ $=(2sin((4pi)/7)cos((4pi)/7))/(8sin(pi/7))$

$= \frac{sin (\frac{8 π}{7})}{8 sin (\frac{π}{7})}$ $=sin((8pi)/7)/(8sin(pi/7))$

$= \frac{sin (π + \frac{π}{7})}{8 sin (\frac{π}{7})}$ $=sin(pi+pi/7)/(8sin(pi/7))$

$= - \frac{sin (\frac{π}{7})}{8 sin (\frac{π}{7})} = - \frac{1}{8}$ $=-sin(pi/7)/(8sin(pi/7))=-1/8$

So the whole LHS

$= - \frac{- \frac{1}{2}}{- \frac{1}{8}} = - 4 = R H S$ $=-(-1/2)/(-1/8)=-4=RHS$

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Question #acd9c

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