Solution :
\(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\)
\(\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\)
\(\tan(\frac{\pi}{4}) = 1\)
Résultat : \(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} – 1 = \sqrt{3} – 1\)

Solution :
\(\sin^2x + \cos^2x = 1 \Rightarrow \cos x = \sqrt{1 – (\frac{3}{5})^2} = \frac{4}{5}\) (car cos positif sur \([0, \frac{\pi}{2}]\))
\(\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{4/5} = \frac{3}{4}\)

Solution :
\(\sin(a-b) = \sin a \cos b – \cos a \sin b\)
\(\sin(\frac{\pi}{12}) = \sin(\frac{\pi}{3})\cos(\frac{\pi}{4}) – \cos(\frac{\pi}{3})\sin(\frac{\pi}{4})\)
\(= \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2} – \frac{1}{2} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{6} – \sqrt{2}}{4}\)

Solution :
On reconnaît \(\sin(a-b) = \sin a \cos b – \cos a \sin b\) avec \(a = 2x\) et \(b = x\)
Donc \(\sin(2x) \cos x – \cos(2x) \sin x = \sin(2x – x) = \sin x\)

Solution :
\(\sin x = \frac{1}{2} = \sin(\frac{\pi}{6})\)
Solutions générales :
\(x = \frac{\pi}{6} + 2k\pi\) ou \(x = \frac{5\pi}{6} + 2k\pi\), \(k \in \mathbb{Z}\)

Solution :
Poser \(X = \cos x\) : \(2X^2 – X – 1 = 0\)
Solutions : \(X = 1\) ou \(X = -\frac{1}{2}\)
• \(\cos x = 1 \Rightarrow x = 0\) ou \(2\pi\)
• \(\cos x = -\frac{1}{2} \Rightarrow x = \frac{2\pi}{3}\) ou \(\frac{4\pi}{3}\)
Solutions dans \([0, 2\pi]\) : \(0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi\)

Solution :
\(\cos^4 x – \sin^4 x = (\cos^2 x – \sin^2 x)(\cos^2 x + \sin^2 x)\)
\(= (\cos^2 x – \sin^2 x) \times 1 = \cos(2x)\)

Solution :
• \(\cos(\frac{\pi}{6}) = \frac{AB}{BC} \Rightarrow BC = \frac{5}{\frac{\sqrt{3}}{2}} = \frac{10\sqrt{3}}{3}\)
• \(\tan(\frac{\pi}{6}) = \frac{AC}{AB} \Rightarrow AC = 5 \times \frac{\sqrt{3}}{3} = \frac{5\sqrt{3}}{3}\)

Solution :
Méthode : diviser par 2 pour reconnaître une formule d’addition
\(\frac{1}{2}\sin x + \frac{\sqrt{3}}{2}\cos x = \frac{1}{2}\)
\(\sin(x + \frac{\pi}{3}) = \frac{1}{2}\)
Solutions : \(x + \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi\) ou \(x + \frac{\pi}{3} = \frac{5\pi}{6} + 2k\pi\)
Donc \(x = -\frac{\pi}{6} + 2k\pi\) ou \(x = \frac{\pi}{2} + 2k\pi\), \(k \in \mathbb{Z}\)

Solution :
On multiplie en croix : \((1 + \sin x)(1 – \sin x) = \cos^2 x\)
\(1 – \sin^2 x = \cos^2 x\) ce qui est vrai car \(\sin^2 x + \cos^2 x = 1\)