If sin⁡5θ=asin5⁡θ+bsin3⁡θ+csin⁡θ+d, then

Sin 5θ=sin⁡(2θ+3θ)=sin⁡2θcos⁡3θ+cos⁡2θsin⁡3θ=2sin⁡θcos⁡θ×cos⁡θ4cos2⁡θ−3+1−2sin2⁡θ3sin⁡θ−4sin3⁡θ =2sin⁡θ1−sin2⁡θ1−4sin2⁡θ+1−2sin2⁡θ3sin⁡θ−4sin3⁡θ =16sin5⁡θ−20sin3⁡θ+5sin⁡θ So a=16,b=−20,c=5,d=0⇒5a+3b−4c=0