$sin(\alpha+2\alpha)=sin2\alpha.cos\alpha+sin\alpha.cos2\alpha$ deyip
$cos2\alpha=1-2sin^2\alpha$ derız
üçgen çizip $cos\alpha=\frac{\sqrt{1-a^2}}{1}$ oldugunu goruruz yerlerıne yazarsak
$sin(\alpha+2\alpha)=2sin\alpha.cos^2\alpha+sin\alpha.cos2\alpha$
$sin(\alpha+2\alpha)=sin\alpha(2cos^2\alpha+cos2\alpha)$
$sin(\alpha+2\alpha)=a(2(1-a^2)+(1-2a^2))=a[2-2a^2+1-2a^2]=3a-4a^3$ olur