Ingat bahwa $\cos72\degree$ dapat ditulis menjadi $\cos\left(90\degree-18\degree\right)$

Hubungan pada kuadran I, $\cos\left(90\degree-\theta\right)=\sin\theta$. Dengan demikian,

$\cos\left(90\degree-18\degree\right)=\sin18\degree$

Misalkan, $A=18\degree$ maka $5A=90\degree$

$5A=90\degree$

$2A+3A=90\degree$

$2A=90\degree-3A$

$\sin2A=\sin\left(90\degree-3A\right)$

Hubungan pada kuadran I, $\sin\left(90\degree-\theta\right)=\cos\theta$

$\sin2A=\cos3A$

$\sin2A=\cos\left(2A+A\right)$

Karena $\sin2A=2\sin A\cos\ A$ dan

$\cos\left(\alpha+\beta\right)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ maka

$2\sin A\cos A=\cos2A\cos A-\sin2A\sin A$

$2\sin A\cos A=\cos2A\cos A-\left(2\sin A\cos A\right)\sin A$

$2\sin A\cos A=\cos2A\cos A-2\sin^2A\cos A$

Karena $\sin^2A=1-\cos^2A$ maka

$2\sin A\cos A=\cos2A\cos A-2\left(1-\cos^2A\right)\cos A$

$2\sin A\cos A=\cos2A\cos A-2\cos A+2\cos^3A$

Karena $\cos2A=2\cos^2A-1$ maka

$2\sin A\cos A=\left(2\cos^2A-1\right)\cos A-2\cos A+2\cos^3A$

$2\sin A\cos A=2\cos^3A-\cos A-2\cos A+2\cos^3A$

$2\sin A\cos A=4\cos^3A-3\cos A$

$2\sin A\cos A-4\cos^3A+3\cos A=0$

$\cos A\left(2\sin A-4\cos^2A+3\right)=0$

Setiap ruas dikali $\frac{1}{\cos A}$

$2\sin A-4\cos^2A+3=0$

Karena $\cos^2A=1-\sin^2A$ maka

$2\sin A-4\left(1-\sin^2A\right)+3=0$

$2\sin A-4+4\sin^2A+3=0$

$4\sin^2A+2\sin A-1=0$

Mencari akar dari persamaan kuadrat menggunakan rumus $ABC$

Jika diketahui persamaan $ax^2+bx+c=0$ maka rumus $ABC$

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Karena $a=4,b=2,c=-1$ maka

$\sin A=\frac{-2\pm\sqrt{2^2-4\left(4\right)\left(-1\right)}}{2\left(4\right)}$

$=\frac{-2\pm\sqrt{4+16}}{8}$

$=\frac{-2\pm\sqrt{20}}{8}$

$=\frac{-2\pm2\sqrt{5}}{8}$

$\sin A=\frac{-1\pm\sqrt{5}}{4}$

Karena sudut berada di kuadran I, maka sinus bernilai positif

$$ $\sin A=\frac{\sqrt{5}-1}{4}$

Jadi, $\cos72\degree=\sin18\degree=\frac{\sqrt{5}-1}{4}$