Question #0912d

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Question #0912d

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Answer 1 · 2016-12-20T14:37:40

$L H S = {sin}^{2} θ + {sin}^{2} (θ + \frac{π}{3}) + {sin}^{2} (θ - \frac{π}{3})$ $LHS=sin ^2 theta + sin^2 (theta+pi/3) + sin^2 (theta-pi/3)$

$= \frac{1}{2} (2 {sin}^{2} θ + 2 {sin}^{2} (θ + \frac{π}{3}) + 2 {sin}^{2} (θ - \frac{π}{3})$ $=1/2(2sin ^2 theta + 2sin^2 (theta+pi/3) + 2sin^2 (theta-pi/3)$

$= \frac{1}{2} (1 - cos 2 θ + 1 - cos 2 (θ + \frac{π}{3}) + 1 - cos 2 (θ - \frac{π}{3})$ $=1/2(1-cos2 theta + 1-cos2 (theta+pi/3) + 1-cos2 (theta-pi/3)$

$= \frac{1}{2} (3 - cos 2 θ - (cos 2 (θ + \frac{π}{3}) + cos 2 (θ - \frac{π}{3}))$ $=1/2(3-cos2 theta -(cos2 (theta+pi/3) + cos2 (theta-pi/3))$

$= \frac{1}{2} (3 - cos 2 θ - 2 cos 2 θ cos (\frac{2 π}{3}))$ $=1/2(3-cos2 theta -2cos2 thetacos((2pi)/3))$

$= \frac{1}{2} (3 - cos 2 θ - 2 cos 2 θ \times (- \frac{1}{2}))$ $=1/2(3-cos2 theta -2cos2 thetaxx(-1/2))$

$= \frac{3}{2} = R H S$ $=3/2=RHS$

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Question #0912d

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