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Question #33ab5

1 Answer

Dec 20, 2016

$R H S = sin 7 θ + sin 4 θ - sin 2 θ + sin θ$ $RHS=sin7theta+sin4theta-sin2theta+sintheta$

$= (sin 7 θ + sin θ) + (sin 4 θ - sin 2 θ)$ $=(sin7theta+sintheta)+(sin4theta-sin2theta)$

$= (2 sin 4 θ cos 3 θ) + (2 cos 3 θ sin θ)$ $=(2sin4thetacos3theta)+(2cos3thetasintheta)$

$= 2 (sin 4 θ + sin θ) cos 3 θ$ $=2(sin4theta+sintheta)cos3theta$

$= 4 sin (\frac{5 θ}{2}) cos (\frac{3 θ}{2}) cos 3 θ = L H S$ $=4sin((5theta)/2)cos((3theta)/2)cos3theta=LHS$

Proved

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