導出
\[\begin{aligned} \sin3\theta=&\sin(2\theta+\theta)\\ =&\sin2\theta\cos\theta+\cos2\theta\sin\theta\\ =&2\sin\theta\cos^2\theta+(1-2\sin^2\theta)\sin\theta\\ =&2\sin\theta(1-\sin^2\theta)+\sin\theta-2\sin^3\theta\\ =&2\sin\theta-2\sin^3\theta+\sin\theta-2\sin^3\theta\\ =&3\sin\theta-4\sin^3\theta \end{aligned}\]
\[\begin{aligned} \cos3\theta=&\cos(2\theta+\theta)\\ =&\cos2\theta\cos\theta-\sin2\theta\sin\theta\\ =&(2\cos^2\theta-1)\cos\theta-2\sin^2\theta\cos\theta\\ =&2\cos^3\theta-\cos\theta-2(1-\cos^2\theta)\cos\theta\\ =&2\cos^3\theta-\cos\theta-2\cos\theta+2\cos^3\theta\\ =&4\cos^3\theta-3\cos\theta\\ \end{aligned}\]
\[\begin{aligned} \cos3\theta=&\sin\left(\dfrac92\pi-3\theta\right)\\ =&3\sin\left(\dfrac32\pi-\theta\right)-4\sin^3\left(\dfrac32\pi-\theta\right)\\ =&-3\cos\theta+4\cos^3\theta \end{aligned}\]別の導出のやり方
\[\sin3\theta+\sin\theta=2\sin2\theta\cos\theta\] \[\begin{aligned} \sin3\theta=&2\sin2\theta\cos\theta-\sin\theta\\ =&4\sin\theta\cos^2\theta-\sin\theta\\ =&4\sin\theta(1-\sin^2\theta)-\sin\theta\\ =&4\sin\theta-4\sin^3\theta-\sin\theta\\ =&3\sin\theta-4\sin^3\theta\\ \end{aligned}\]
\[\cos3\theta+\cos\theta=2\cos2\theta\cos\theta\] \[\begin{aligned} \cos3\theta=&2\cos2\theta\cos\theta-\cos\theta\\ =&2(2\cos^2\theta-1)\cos\theta-\cos\theta\\ =&4\cos^3\theta-2\cos\theta-\cos\theta\\ =&4\cos^3\theta-3\cos\theta\\ \end{aligned}\]