1°

${\displaystyle {\begin{aligned}{\frac {\sin a+\sin b+\sin c}{\sin a+\sin b-\sin c}}&={\frac {\sin a+\sin b+\sin(a+b)}{\sin a+\sin b-\sin(a+b)}}\\&={\frac {2\sin((a+b)/2))\left(\cos((a-b)/2))+\cos((a+b)/2))\right)}{2\sin((a+b)/2))\left(\cos((a-b)/2))-\cos((a+b)/2))\right)}}\\&={\frac {\cos(a/2)\cos(b/2)}{\sin(a/2)\sin(b/2)}}\\&=\cot(a/2)\cot(b/2)\end{aligned}}}$

2°

${\displaystyle {\begin{aligned}1+\cos a+\cos b-\cos c&=\cos a+\cos b+1+\cos(a+b)\\&=2\cos((a+b)/2)\left(\cos((a-b)/2)+\cos((a+b)/2)\right)\\&=2\cos((\pi -c)/2)\times 2\cos a\cos b\\&=4\sin {\frac {c}{2}}\cos a\cos b\end{aligned}}}$

${\displaystyle {\begin{aligned}S&=2\left(\sin ^{4}{\frac {\pi }{8}}+\cos ^{4}{\frac {\pi }{8}}\right)\\&=2\left(\left(\sin ^{2}{\frac {\pi }{8}}+\cos ^{2}{\frac {\pi }{8}}\right)^{2}-2\sin ^{2}{\frac {\pi }{8}}\cos ^{2}{\frac {\pi }{8}}\right)\\&=2-\sin ^{2}{\frac {\pi }{4}}\\&={\frac {3}{2}}.\end{aligned}}}$

${\displaystyle \cos ^{4}x=\left({\frac {1+\cos 2x}{2}}\right)^{2}={\frac {1+2\cos 2x+\cos ^{2}2x}{4}}={\frac {1+2\cos 2x+{\frac {1+\cos 4x}{2}}}{4}}={\frac {3}{8}}+{\frac {\cos 2x}{2}}+{\frac {\cos 4x}{8}}}$.

${\displaystyle {\frac {S}{2}}=\cos ^{4}{\frac {\pi }{8}}+\cos ^{4}{\frac {3\pi }{8}}={\frac {3}{4}}+{\frac {\cos {\frac {\pi }{4}}+\cos {\frac {3\pi }{4}}}{2}}+{\frac {\cos {\frac {\pi }{2}}+\cos {\frac {3\pi }{2}}}{8}}={\frac {3}{4}}}$ donc ${\displaystyle S={\frac {3}{2}}}$.

1°  ${\displaystyle \sin a+2\sin 2a+\sin 3a=2\sin 2a\left(1+\cos a\right)=4\sin 2a\cos ^{2}{\frac {a}{2}}}$.

2°  ${\displaystyle \cos a+2\cos 2a+\cos 3a=2\cos 2a\left(1+\cos a\right)=4\cos 2a\cos ^{2}{\frac {a}{2}}}$.

3°  ${\displaystyle \sin ^{2}a+\sin ^{2}b-\sin ^{2}(a+b)=\sin ^{2}a+\sin ^{2}b-\left(\sin a\cos b+\sin b\cos a\right)^{2}}$

${\displaystyle =\sin ^{2}a\left(1-\cos ^{2}b\right)+\sin ^{2}b\left(1-\cos ^{2}a\right)-2\sin a\cos a\sin b\cos b=2\sin a\sin b\left(\sin a\sin b-\cos a\cos b\right)}$

${\displaystyle =-2\sin a\sin b\cos \left(a+b\right)}$.

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

5°  ${\displaystyle \cos ^{2}a-\cos ^{2}b=\left(\cos a+\cos b\right)\left(\cos a-\cos b\right)=-4\cos {\frac {a+b}{2}}\cos {\frac {a-b}{2}}\sin {\frac {a+b}{2}}\sin {\frac {a-b}{2}}=-\sin \left(a+b\right)\sin \left(a-b\right)}$.

1°  ${\displaystyle 1-\sin ^{2}x-\sin ^{2}y=\cos ^{2}x-\cos ^{2}\left({\frac {\pi }{2}}-y\right)=-\sin \left(x+{\frac {\pi }{2}}-y\right)\sin \left(x-{\frac {\pi }{2}}+y\right)}$ (cf. fin de l'exercice précédent).

2°  ${\displaystyle \cos 2x-\cos ^{2}x=\cos ^{2}x-1=-\sin ^{2}x}$.

3°  ${\displaystyle \sin x+\sin 2x=2\sin x\left({\frac {1}{2}}+\cos x\right)=2\sin x\left(\cos {\frac {\pi }{3}}+\cos x\right)=4\sin x\cos {\frac {{\frac {\pi }{3}}+x}{2}}\cos {\frac {{\frac {\pi }{3}}-x}{2}}}$.

4°  ${\displaystyle 1+\tan x\tan 2x=1+{\frac {2\tan x\cos x\sin x}{\cos 2x}}={\frac {\cos 2x+2\sin ^{2}x}{\cos 2x}}={\frac {1}{\cos 2x}}}$.

5°  ${\displaystyle 1+\cos 2x+2\cos x=2\cos x\left(\cos x+1\right)=4\cos x\cos ^{2}{\frac {x}{2}}}$.

1°  La partie réelle de ${\displaystyle 1-\left(\cos t+\mathrm {i} \sin t\right)^{18}}$ est égale à ${\displaystyle 1-\sum _{k=0}^{9}{\binom {18}{2k}}(-1)^{k}\sin ^{2k}t\left(1-\sin ^{2}t\right)^{9-k}=\sin ^{2}t\,P(\sin t)}$, où ${\displaystyle P}$ est un polynôme de degré 16 à 8 racines doubles : ${\displaystyle \pm \sin {\frac {\pi }{9}},\pm \sin {\frac {2\pi }{9}},\pm \sin {\frac {3\pi }{9}},\pm \sin {\frac {4\pi }{9}}}$. Le coefficient dominant de ${\displaystyle P}$ est ${\displaystyle \sum _{k=0}^{9}{\binom {18}{2k}}=2^{17}}$ et son terme constant est ${\displaystyle -{\binom {18}{0}}(-1)^{0}(-9)-{\binom {18}{2}}(-1)^{1}=9+{\frac {18\times 17}{2}}=9\times 18}$, donc ${\displaystyle S={\sqrt[{4}]{\frac {9\times 18}{2^{17}}}}={\frac {3}{16}}}$.

2°  La partie réelle de ${\displaystyle 1-\left(\cos t+\mathrm {i} \sin t\right)^{18}}$ est aussi égale à ${\displaystyle 1-\sum _{k=0}^{9}{\binom {18}{2k}}(-1)^{k}\left(1-\cos ^{2}t\right)^{k}\cos ^{18-2k}t=\left(1-\cos ^{2}t\right)\,Q(\cos t)}$, où ${\displaystyle Q}$ est un polynôme de degré 16 à 8 racines doubles : ${\displaystyle \pm \cos {\frac {\pi }{9}},\pm \cos {\frac {2\pi }{9}},\pm \cos {\frac {3\pi }{9}},\pm \cos {\frac {4\pi }{9}}}$. Le coefficient dominant de ${\displaystyle Q}$ est ${\displaystyle 2^{17}}$ et son terme constant est ${\displaystyle 2}$, donc ${\displaystyle C={\sqrt[{4}]{\frac {2}{2^{17}}}}={\frac {1}{16}}}$.