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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+sinA4 sinA cosA2cos3A2
(v) cos20cos100+cos100 cos140cos200°=34

(vi) sinθ2sin7θ2+sin3θ2sin11θ2=sin2θ sin5θ
(vii) sinθ2cos3θ cos=9θ2=sin7θ sin8θ


Solution

(i) cos3A+cosSA+cos7A+cos15A 4cos4A cosSA cos6A
We have,
LHS=cos3A+cos5A+cos7A+cos15A=[cos5A+cos3A]+[cos15A+cos7A]=[2cos(5A+3A)2cos(5A3A)2]+[2cos(15A+7A)2cos(15A7A)2]=2cos4A cosA+2cos11A cos4A=2cos4A[cos A+cos11A]=2cos4A[2cos(11A+A)2cos(11AA)2]=4cosA[cos6A cos5A]=4cos 4Acos 5Acos6A=RHScos3A+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(3AA2)]+[2cos(7A+5A2)cos(7A5A2)]=2cos2AcosA+2cos6AcosA=2cosA[cos2A+cos6A]=2cosA[cos6A+cos2A]=2cosA[2cos(6A+2A2)cos(6A2A2)]=4coaA[cis4A cos2A]cosA+cos3A+cos5A+cos7A=4cosAcos2Acos4A.

(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(5A4A2)]=2sin3A2cossA2+2sin9A2cosA2=2cosA2[sin3A2+sin9A2]=2cosA2[sin9A2+2sin3A2]=[2cosA22sin{12(9A2+3A2)}cos{12(9A23A2)}]=4cosA2[sin12A4cos6A4]=4cosA2sin3Acos3A2=4cosA2cos3A2sin3A=RHSsinA+sin2A+sin4A+sin5A+=4cosA2cos3A2sin3A

(iv) We have,
sin3A+sin2A+sinA4 sinA cosA2cos3A2=sin3A+sin2AsinA=sin3Asin2Asin2A=2sin(3AA2)cos(3A+A2)+sin2A=2sinA cos2A+sin2A=2sinA cos2A+2sinA cosA=2sinA[cos2A+cosA]=2sinA[2cos(2A+A2)cos(2AA2)]=4sinA cos3A2cosA2=4sinA cosA2cos3A2=RHSsin3A+sin2Asin A=4sin AcosA2cos3A2

(v) cos20cos100+cos100 cos140cos200°=34LHS=cos20cos100+cos100 cos140cos200°=12[2cos100 cos20+2cos140 cos1002cos200 cos140]=12[cos(100+20)+cos(10020)+cos(140+100)+cos(140100){cos(200+140)+cos(200140}]=12[cos120+cos80+cos240+cos40cos340cos60]=12[cos(90+30)+cos80+cos40cos(80+60)cos(36020)12]=12[sin30+2cos(80+402)cos(80402)cos60cos2012]=12[12+2cos60 cos2012cos2012]=12[32+2×12cos20cos20]=12[32+cos20cos20]=12[32+0]=34=RHScos20 cos100+cos100 cos140cos140 cos200=34

(vi) sinθ2sin7θ2+sin3θ2sin11θ2=sin2θ sin5θ
We have,
LHS=sinθ2sin7θ2+sin3θ2sin11θ2=12[2sin7θ2sinθ2+2sin11θ2sin3θ2]=12[cos(7θ2θ2)cos(7θ2θ2)+cos(11θ23θ2)cos(11θ23θ2)]=12[cos6θ2cos8θ2+cos8θ2cos14θ2]=12[cos3θcos4θ+cos4θcos7θ]=12[cos3θcos7θ]=12[cos7θcos3θ]=12[2sin(7θ+3θ2)sin(7θ3θ2)]=sin10θ2sin4θ2=sin5θ sin2θ=sin2θ sin5θ=RHSsinθ2 sin7θ2+sin3θ2 sin11θ2=sin2θ sin5θ

(vii) cosθ cosθ2cos3θ cos=9θ2=sin7θ sin8θLHS=cosθ cosθ2cos3θ cos9θ2=12[2cosθ.cosθ22cos3θ.cos9θ2]=12[cos(θ+θ2)+cos(θθ2)cos(3θ+9θ2)cos(3θ9θ2)]=12(cos3θ2+cosθ2cos15θ2cos3θ2)=12[cosθ2cos15θ2]=12[2sin(θ+15θ2)sin(θ15θ2)][cosxcosy=2sinx+y2.sinxy2]=+(sin8θ.sin7θ)=RHSLHS=RHS


Mathematics
RD Sharma
Standard XI

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