Question #9351d

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Question #9351d

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Answer 1 · 2016-11-10T05:08:04

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

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

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

using formula ${cos}^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ $cos^2theta=1/2(1+cos2theta)$

$= \frac{1}{2} (1 + cos (\frac{2 π}{7})) + \frac{1}{2} (1 + cos (\frac{4 π}{7})) + \frac{1}{2} (1 + cos (\frac{8 π}{7}))$ $=1/2(1+cos((2pi)/7))+1/2(1+cos((4pi)/7))+ 1/2(1+cos((8pi)/7))$

$= \frac{3}{2} + \frac{1}{2} (cos (2 \frac{π}{7}) + cos (4 \frac{π}{7}) + cos (\frac{8 π}{7}))$ $=3/2+1/2(cos(2pi/7)+cos(4pi/7)+cos((8pi)/7))$

$= \frac{3}{2} + \frac{1}{4 sin (\frac{π}{7})} (2 sin (\frac{π}{7}) cos (\frac{2 π}{7}) + 2 sin (\frac{π}{7}) cos (\frac{4 π}{7}) + 2 sin (\frac{π}{7}) cos (\frac{8 π}{7}))$ $=3/2+1/(4sin(pi/7)) (2sin(pi/7)cos((2pi)/7)+2sin(pi/7)cos((4pi)/7)+2sin(pi/7)cos((8pi)/7))$

$=3/2+1/(4sin(pi/7)) (cancel(sin((3pi)/7))-sin(pi/7)+sin((5pi)/7)-cancel(sin((3pi)/7))+sin((9pi)/7)-sin((7pi)/7))$

$=3/2+1/(4sin(pi/7)) (-sin(pi/7)+sin(pi-(2pi)/7)+sin(pi+(2pi)/7)-0)$

$=3/2+1/(4sin(pi/7)) (-sin(pi/7)+cancel(sin((2pi)/7))-cancel(sin(2pi/7)))$

$=3/2+1/(4cancel(sin(pi/7))) (-cancel(sin(pi/7)))$

$=3/2-1/4=(6-1)/4=5/4$

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Question #9351d

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