(a)We know from our knowledge on Complex Numbers that,cos(nθ)+isin(nθ)cos(7θ)+isin(7θ)=(cosθ+isinθ)n=(cosθ+isinθ)7=cos7θ+7cos6θ(isinθ)+21cos5θ(isinθ)2+35cos4θ(isinθ)3+35cos3θ(isinθ)4+21cos2θ(isinθ)5+7cosθ(isinθ)6+(isinθ)7=cos7θ+7icos6θsinθ−21cos5θsin2θ−35icos4θsin3θ+35cos3θsin4θ+21icos2θsin5θ−7cosθsin6θ−isin7θComparing the real partscos(7θ)=cos7θ−21cos5θsin2θ+35cos3θsin4θ−7cosθsin6θ(b)cos(nθ)+isin(nθ)cos(4θ)+isin(4θ)=(cosθ+isinθ)n=(cosθ+isinθ)4=cos4θ+4cos3θ(isinθ)+6cos2θ(−sin2θ)+4cos3θ(−isin3θ)+sin4θComparing the real partscos(4θ)=cos4θ+sin4θ−6cos2θsin2θ(2)Alsocos(3θ)+isin(3θ)=(cosθ+sinθ)3=cos3θ+3cos2θ(isinθ)+3cosθ(−sin2θ)−isin3θ(3)Comparing the imaginary partssin(3θ)=3cos2θsinθ−sin3θTo derive the value ofcos(4θ)sin(3θ),Multiply the result inequation(3)by the result in equation(2)∴cos(4θ)sin(3θ)=(cos4θ+sin4θ−6cos2θsin2θ)(3cos2θsinθ−sin3θ)=3cos6θsinθ+3cos2θsin5θ−18cos4θsin3θ−cos4θsin3θ−sin7θ+6cos2θsin5θ=3cos6θsinθ+9cos2θsin5θ−19cos4θsin3θ−sin7θ⟹cos(4θ)sin(3θ)=3cos6θsinθ+9cos2θsin5θ−19cos4θsin3θ−sin7θ(c)(i)Prove thatsin(2A)+sin(2B)+sin(2C)=4sinAsinBsinCSince the figure is a planar triangleA+B+C=πIt is known that,sinP+sinQ=2sin(2P+Q)cos(2P−Q)sin(2A)+sin(2B)+sin(2C)=2sin(A+B)cos(A−B)+sin(2C)=2sin(π−C)cos(A−B)+2sin(C)cos(C)=2sin(C)cos(A−B)+2sin(C)cos(C)=2sin(C)(cos(A−B)+cos(C))It is also known that,cosP+cosQ=2cos(2P+Q)cos(2P−Q)⟹sin(2A)+sin(2B)+sin(2C)=2sin(C)(2cos(2A+C−B)cos(2A−B−C))=4sinCcos(2π−B)cos(2π−A)=4sinCsinBsinA=4sinAsinBsinC(ii)cos(2A)+cos(2B)+cos(2C)=2cos(22(A+B))cos(22(A−B))+cos(2C)=2cos(π−C)cos(A−B)+2cos2C−1=2cosC(cosC−cos(A−B))−1It is also known that,cosP−cosQ=2sin(2P+Q)sin(2P−Q)cos(2A)+cos(2B)+cos(2C)=2cosC(2sin(2A+C−B)sin(2C−A+B))−1=2cosC(2sin(2π−B)sin(2π−A))−1=2cosC(2cosB×2cosA)−1=4cosAcosBcosC−1
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