a) Es ist ${\displaystyle {}\sin \left(z+{\frac {\pi }{2}}\right)=\cos z}$ nach Fakt  (3). Daher ist

${\displaystyle {}\sin \left({\frac {\pi }{4}}\right)=\sin \left(-{\frac {\pi }{4}}+{\frac {\pi }{2}}\right)=\cos \left(-{\frac {\pi }{4}}\right)=\cos \left({\frac {\pi }{4}}\right)\,,}$

da Kosinus eine gerade Funktion ist. Aus

${\displaystyle {}1=\sin ^{2}{\left({\frac {\pi }{4}}\right)}+\cos ^{2}{\left({\frac {\pi }{4}}\right)}=2\sin ^{2}{\left({\frac {\pi }{4}}\right)}\,}$

ergibt sich

${\displaystyle {}\sin ^{2}{\left({\frac {\pi }{4}}\right)}={\frac {1}{2}}\,.}$

Da ${\displaystyle {}\sin {\frac {\pi }{4}}>0}$ ist, ist

${\displaystyle {}\sin {\frac {\pi }{4}}={\frac {1}{\sqrt {2}}}\,.}$

b) Nach den Additionstheoremen für Sinus und Kosinus ist

${\displaystyle {}{\begin{aligned}\sin \left(3\alpha \right)&=\cos \alpha \sin \left(2\alpha \right)+\cos \left(2\alpha \right)\sin \alpha \\&=\cos \alpha {\left(2\cos \alpha \sin \alpha \right)}+{\left(\cos ^{2}\alpha -\sin ^{2}\alpha \right)}\sin \alpha \\&=\sin \alpha {\left(3\cos ^{2}\alpha -\sin ^{2}\alpha \right)}\\&=\sin \alpha {\left(4\cos ^{2}\alpha -1\right)}.\end{aligned}}}$

Für ${\displaystyle {}\alpha ={\frac {\pi }{3}}}$ ist also

${\displaystyle {}0=\sin \pi =\sin {\frac {\pi }{3}}{\left(4\cos ^{2}{\frac {\pi }{3}}-1\right)}\,.}$

Wegen

${\displaystyle {}\sin {\frac {\pi }{3}}\neq 0\,}$

ist somit

${\displaystyle {}0=4\cos ^{2}{\frac {\pi }{3}}-1\,,}$

woraus sich

${\displaystyle {}\cos ^{2}{\frac {\pi }{3}}={\frac {1}{4}}\,}$

ergibt. Da ${\displaystyle {}\cos {\frac {\pi }{3}}}$ positiv ist, folgt

${\displaystyle {}\cos {\frac {\pi }{3}}={\frac {1}{2}}\,.}$

c) Aus

${\displaystyle {}1=\cos ^{2}{\frac {\pi }{3}}+\sin ^{2}{\frac {\pi }{3}}={\frac {1}{4}}+\sin ^{2}{\frac {\pi }{3}}\,}$

folgt

${\displaystyle {}\sin ^{2}{\frac {\pi }{3}}={\frac {3}{4}}\,,}$

woraus sich wegen der Positivität von ${\displaystyle {}\sin {\frac {\pi }{3}}}$ schließlich

${\displaystyle {}\sin {\frac {\pi }{3}}={\frac {\sqrt {3}}{2}}\,}$