cos3⁡xsin⁡2x=∑x=0n arsin⁡(rx)∀x∈R then

cos3⁡xsin⁡2x=cos2⁡xcos⁡xsin⁡2x=1+cos⁡2x22sin⁡2xcos⁡x2=14(1+cos⁡2x)(sin⁡3x+sin⁡x)=14sin⁡3x+sin⁡x+12(2sin⁡3xcos⁡2x)+12(2cos⁡2xsin⁡x)=14sin⁡3x+sin⁡x+12(sin⁡5x+sin⁡x)+12(sin⁡3x−sin⁡x)=14sin⁡x+32sin⁡3x+12sin⁡5x⇒ a1=1/4,a3=3/8,n=5