Question #28025

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Question #28025

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Answer 1 · 2016-12-18T17:24:39

$L H S = sin (\frac{π}{9}) sin (\frac{2 π}{9}) sin (\frac{3 π}{9}) sin (\frac{4 π}{9})$ $LHS=sin (pi/9) sin ((2pi)/9 )sin ((3pi)/9 )sin ((4pi)/9)$

$= sin (\frac{π}{9}) sin (\frac{2 π}{9}) sin (\frac{π}{3}) sin (\frac{4 π}{9})$ $=sin (pi/9) sin ((2pi)/9 )sin (pi/3 )sin ((4pi)/9)$

$= \frac{\sqrt{3}}{2} sin (\frac{π}{9}) sin (\frac{2 π}{9}) sin (\frac{4 π}{9})$ $=sqrt3/2sin (pi/9) sin ((2pi)/9 )sin ((4pi)/9)$

$= \frac{\sqrt{3}}{4} (2 sin (\frac{π}{9}) sin (\frac{4 π}{9})) sin (\frac{2 π}{9})$ $=sqrt3/4(2sin (pi/9) sin ((4pi)/9 ))sin ((2pi)/9)$

$= \frac{\sqrt{3}}{4} (cos (\frac{4 π - π}{9}) - cos (\frac{4 π + π}{9})) sin (\frac{2 π}{9})$ $=sqrt3/4(cos ((4pi-pi)/9) -cos ((4pi+pi)/9 ))sin ((2pi)/9)$

$= \frac{\sqrt{3}}{4} (cos (\frac{π}{3}) - cos (\frac{5 π}{9})) sin (\frac{2 π}{9})$ $=sqrt3/4(cos (pi/3) -cos ((5pi)/9 ))sin ((2pi)/9)$

$= \frac{\sqrt{3}}{4} (\frac{1}{2} - cos (\frac{5 π}{9})) sin (\frac{2 π}{9})$ $=sqrt3/4(1/2 -cos ((5pi)/9 ))sin ((2pi)/9)$

$= \frac{\sqrt{3}}{8} (sin (\frac{2 π}{9}) - 2 cos (\frac{5 π}{9}) sin (\frac{2 π}{9}))$ $=sqrt3/8(sin((2pi)/9) -2cos ((5pi)/9 )sin ((2pi)/9) )$

$= \frac{\sqrt{3}}{8} (sin (\frac{2 π}{9}) - (sin (\frac{7 π}{9}) - sin (\frac{3 π}{9}))$ $=sqrt3/8(sin((2pi)/9)-(sin((7pi)/9) -sin((3pi)/9 ))$

$= \frac{\sqrt{3}}{8} (sin (\frac{2 π}{9}) - (sin (π - \frac{2 π}{9}) - sin (\frac{π}{3}))$ $=sqrt3/8(sin((2pi)/9)-(sin(pi-(2pi)/9) -sin(pi/3 ))$

$=sqrt3/8(cancelsin((2pi)/9)-cancelsin((2pi)/9) +sqrt3/2)$

$=3/16=RHS$

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Question #28025

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