$t=\frac{7}{8}\text{π}+k\cdot 2\text{π}$ of $t=1\frac{7}{8}\text{π}+k\cdot 2\text{π}$, met $k$ geheel.

$\mathrm{‐}\sin (t)$ ; $\mathrm{‐}\sin (t)$ ; $\sin (t)$ ; $\mathrm{‐}\cos (t)$

${\sin }^2(2x)+{\cos }^2(2x)=1$ (vul in formule 9 voor $t=2x$ in.)
${(2\sin (x\left)\right)}^2+{(2\cos (x\left)\right)}^2=4{\sin }^2(x)+4{\cos }^2(x)=4$.

${\sin }^2(x+\frac{1}{2}\text{π})={\cos }^2(x)$, volgens formule 11.
${\sin }^2(x+1\frac{1}{2}\text{π})={\sin }^2(x+\frac{1}{2}\text{π})$, volgens formule 3.
Dus ${\sin }^2(x)+{\sin }^2(x+\frac{1}{2}\text{π})+{\sin }^2(x+\text{π})+{\sin }^2(x+\text{π})+{\sin }^2(x+1\frac{1}{2}\text{π})=$
${\sin }^2(x)+{\cos }^2(x)+{\sin }^2(x)+{\cos }^2(x)=1+1=2$.

$(1-\cos (x))(1+\cos (x))=1-{\cos }^2(x)$ (merkwaardig product), dus we moeten laten zien: ${\sin }^2(x)=1-{\cos }^2(x)$, dat is formule 9 anders geschreven.

$\cos (x+\frac{1}{3}\text{π})=\sin (\frac{1}{2}\text{π}-(x+\frac{1}{3}\text{π}\left)\right)=\sin (\frac{1}{6}\text{π}-x)$ (Eerst formule 7, dan herschrijven).
Dus $\cos (x+\frac{1}{3}\text{π})+\sin (x-\frac{1}{6}\text{π})=0$ volgens formule 2.

${\sin }^2(\frac{5}{14}\text{π)}={\cos }^2(\frac{2}{14}\text{π)}$ en ${\sin }^2(\frac{4}{14}\text{π)}={\cos }^2(\frac{3}{14}\text{π)}$, dus
${\sin }^2(\frac{2}{14}\text{π})+{\sin }^2(\frac{3}{14}\text{π})+{\sin }^2(\frac{4}{14}\text{π})+{\sin }^2(\frac{5}{14}\text{π)}=$
${\sin }^2(\frac{2}{14}\text{π})+{\cos }^2(\frac{2}{14}\text{π})+{\sin }^2(\frac{3}{14}\text{π})+{\cos }^2(\frac{3}{14}\text{π})=2$.