Question

# Prove that: (i) sinα+sinβ+sinγ−sin(α+β+γ)=4sin(α+β2)sin(β+γ2)sin(γ+α2) (ii) cos(A+B+C)+cos(A−B+C)+cos(A+B−C)+cos(−A+B+C)=4cosA cosB cosC

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Solution

## (i) sinα+sinβ+sinγ−sin(α+β+γ)=4sin(α+β2)sin(β+γ2)sin(γ+α2) We have, LHS=sinα+sinβ+sinγ−sin(α+β+γ)=(sinα+sinβ)+[sinγ−sin(α+β+γ)]=2sin(α+β2)cos(α+β2)+2sin(γ−(α+β+γ)2)cos(γ+(α+β+γ)2)=2sin(α+β2)cos(α−β2)+2sin(−α−β2)cos(α+β+2γ2)=2sin(α+β2)cos(α−β2)−2sin(α+β2)cos(α+β+2γ2)=2sin(α+β2)[cos(α−β2)−cos(α+β+2γ2)]=2sin(α+β2)⎡⎢⎣−2sin[α−β2+]sin[α+β+2γ2]2⎤⎥⎦=2sin(α+β2)[−2sin[1α+2γ2×2]sin[−2β−2γ2×2]]=−4(α+β2)[sin(α+γ2)sin[−(β+γ)2]]=4sin(α+β2)sin(α+γ2)sin(β+γ2)=4sin(α+β2)sin(β+γ2)sin(α+γ2)=RHS∴sinα+sinβ+sinγ−sin(α+β+γ)=4sin(α+β2)sin(β+γ2)sin(γ+α2) (ii) cos(A+B+C)+cos(A−B+C)+cos(A+B−C)+cos(−A+B+C)=4cosA cosB cosC We have, LHS=cos(A+B+C)+cos(A−B+C)+cos(A+B−C)+cos(−A+B+C)=2cos{A+B+C+A−B+C2}cos{A+B+C+A+B−C2}+2⎧⎪⎨⎪⎩cos{A+B−C−A−B+C2}cos{A+B−C+A−B−C2}⎫⎪⎬⎪⎭=2cos{2A+2C2}cos{2B2}+2cos{2B2}cos{2A−2C2}=2cos(A+C)cos(B)+2cos(B)cos(A−C)=2cos(B)[cos(A+C)+cos(A−C)]=2cos(B)[2cos(A+C+A−C2)cos(A+C−A+C2)]=2cos(B)[2cosA cosC]=4cosA cosB cosC

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