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Question

# Prove that: (i) cos3A+cosSA+cos7A+cos15A 4cos4A cosSA cos6A (ii) cosA+cos3A+cos5A+cos7A=4cosA cos2A cos4A (iii) sinA+sin2A+sin4A+sin5A 4cosA2cos3Asin3A (iv) sin3A+sin2A+sinA−4 sinA cosA2cos3A2 (v) cos20∘cos100∘+cos100∘ cos140cos200°=−34 (vi) sinθ2sin7θ2+sin3θ2sin11θ2=sin2θ sin5θ (vii) sinθ2−cos3θ cos=9θ2=sin7θ sin8θ

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Solution

## (i) cos3A+cosSA+cos7A+cos15A 4cos4A cosSA cos6A We have, LHS=cos3A+cos5A+cos7A+cos15A=[cos5A+cos3A]+[cos15A+cos7A]=[2cos(5A+3A)2cos(5A−3A)2]+[2cos(15A+7A)2cos(15A−7A)2]=2cos4A cosA+2cos11A cos4A=2cos4A[cos A+cos11A]=2cos4A[2cos(11A+A)2cos(11A−A)2]=4cosA[cos6A cos5A]=4cos 4Acos 5Acos6A=RHS∴cos3A+cos5A+cos7A+cos15A=4cos4A cos5A cos6A (ii) cosA+cos3A+cos5A+cos7A=4cosA cos2A cos4A LHS=cosA+cos3A+cos5A+cos7A=(cos3A+cosA)−(cos7A+cos5A)[2cos(3A+A2)cos(3A−A2)]+[2cos(7A+5A2)cos(7A−5A2)]=2cos2AcosA+2cos6AcosA=2cosA[cos2A+cos6A]=2cosA[cos6A+cos2A]=2cosA[2cos(6A+2A2)cos(6A−2A2)]=4coaA[cis4A cos2A]∴cosA+cos3A+cos5A+cos7A=4cosAcos2‘Acos4A. (iii) sinA+sin2A+sin4A+sin5A 4cosA2cos3Asin3AWehave,LHS=sinA+sin2A+sin4A+sin5A=(sin2A+sinA)+(sin5A+sin4A)=[2cos(2A+A2)cos(2A+A2)]+[2sin(5A+4A2)cos(5A−4A2)]=2sin3A2cossA2+2sin9A2cosA2=2cosA2[sin3A2+sin9A2]=2cosA2[sin9A2+2sin3A2]=[2cosA22sin{12(9A2+3A2)}cos{12(9A2−3A2)}]=4cosA2[sin12A4cos6A4]=4cosA2sin3Acos3A2=4cosA2cos3A2sin3A=RHS∴sinA+sin2A+sin4A+sin5A+=4cosA2cos3A2sin3A (iv) We have, sin3A+sin2A+sinA−4 sinA cosA2cos3A2=sin3A+sin2A−sinA=sin3A−sin2A−sin2A=2sin(3A−A2)cos(3A+A2)+sin2A=2sinA cos2A+sin2A=2sinA cos2A+2sinA cosA=2sinA[cos2A+cosA]=2sinA[2cos(2A+A2)cos(2A−A2)]=4sinA cos3A2cosA2=4sinA cosA2cos3A2=RHS∴sin3A+sin2A−sin A=4sin AcosA2cos3A2 (v) cos20∘cos100∘+cos100∘ cos140cos200°=−34LHS=cos20∘cos100∘+cos100∘ cos140cos200°=12[2cos100∘ cos20∘+2cos140∘ cos100∘−2cos200∘ cos140∘]=12[cos(100∘+20∘)+cos(100∘−20∘)+cos(140∘+100∘)+cos(140∘−100∘)−{cos(200∘+140∘)+cos(200∘−140∘}]=12[cos120∘+cos80∘+cos240∘+cos40∘−cos340∘−cos60∘]=12[cos(90∘+30∘)+cos80∘+cos40∘−cos(80∘+60∘)−cos(360∘−20∘)−12]=12[−sin30∘+2cos(80∘+40∘2)cos(80∘−40∘2)−cos60∘−cos20∘−12]=12[−12+2cos60∘ cos20∘−12−cos20∘−12]=12[−32+2×12cos20∘−cos20∘]=12[−32+cos20∘−cos20∘]=12[−32+0]=−34=RHS∴cos20∘ cos100∘+cos100∘ cos140∘−cos140∘ cos200∘=−34 (vi) sinθ2sin7θ2+sin3θ2sin11θ2=sin2θ sin5θ We have, LHS=sinθ2sin7θ2+sin3θ2sin11θ2=12[2sin7θ2sinθ2+2sin11θ2sin3θ2]=12[cos(7θ2−θ2)−cos(7θ2−θ2)+cos(11θ2−3θ2)−cos(11θ2−3θ2)]=12[cos6θ2−cos8θ2+cos8θ2−cos14θ2]=12[cos3θ−cos4θ+cos4θ−cos7θ]=12[cos3θ−cos7θ]=−12[cos7θ−cos3θ]=−12[−2sin(7θ+3θ2)sin(7θ−3θ2)]=sin10θ2sin4θ2=sin5θ sin2θ=sin2θ sin5θ=RHS∴sinθ2 sin7θ2+sin3θ2 sin11θ2=sin2θ sin5θ (vii) cosθ cosθ2−cos3θ cos=9θ2=sin7θ sin8θLHS=cosθ cosθ2−cos3θ cos9θ2=12[2cosθ.cosθ2−2cos3θ.cos9θ2]=12[cos(θ+θ2)+cos(θ−θ2)−cos(3θ+9θ2)−cos(3θ−9θ2)]=12(cos3θ2+cosθ2−cos15θ2−cos3θ2)=12[cosθ2−cos15θ2]=−12[2sin(θ+15θ2)sin(θ−15θ2)]−[∵cosx−cosy=−2sinx+y2.sinx−y2]=+(sin8θ.sin7θ)=RHS∴LHS=RHS

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