Bonjour
Wendy14,
1) a) Nous savons que : cos(2a) = cos²(a) - sin²(a)
Or sin²(a) + cos²(a) = 1 ==> sin²(a) = 1 - cos²(a)
Donc : cos(2a) = cos²(a) - (1 - cos²(a))
cos(2a) = cos²(a) - 1 + cos²(a)
cos(2a) = 2cos²(a) - 1
Posons [tex]a=\dfrac{\pi}{8}[/tex]
Nous avons ainsi :
[tex]\cos(2\times\dfrac{\pi}{8})=2\cos^2(\dfrac{\pi}{8})-1\\\\\cos(\dfrac{\pi}{4})=2\cos^2(\dfrac{\pi}{8})-1\\\\2\cos^2(\dfrac{\pi}{8})=1+\cos(\dfrac{\pi}{4})\\\\2\cos^2(\dfrac{\pi}{8})=1+\dfrac{\sqrt{2}}{2}\\\\2\cos^2(\dfrac{\pi}{8})=\dfrac{2+\sqrt{2}}{2}\\\\\cos^2(\dfrac{\pi}{8})=\dfrac{2+\sqrt{2}}{4}[/tex]
Or [tex]\cos(\dfrac{\pi}{8})\ \textgreater \ 0\ \ car\ \ 0\ \textless \ \dfrac{\pi}{8}\ \textless \ \dfrac{\pi}{2}[/tex]
Donc
[tex]\cos(\dfrac{\pi}{8})=\sqrt{\dfrac{2+\sqrt{2}}{4}}=\dfrac{\sqrt{2+\sqrt{2}}}{\sqrt{4}}\\\\\\\boxed{\cos(\dfrac{\pi}{8})=\dfrac{\sqrt{2+\sqrt{2}}}{2}}[/tex]
De plus,
[tex]\sin^2(\dfrac{\pi}{8})=1-\cos^2(\dfrac{\pi}{8})\\\\\sin^2(\dfrac{\pi}{8})=1-\dfrac{2+\sqrt{2}}{4}\\\\\sin^2(\dfrac{\pi}{8})=\dfrac{4-2-\sqrt{2}}{4}\\\\\sin^2(\dfrac{\pi}{8})=\dfrac{2-\sqrt{2}}{4}[/tex]
Or [tex]\sin(\dfrac{\pi}{8})\ \textgreater \ 0\ \ car\ \ 0\ \textless \ \dfrac{\pi}{8}\ \textless \ \dfrac{\pi}{2}[/tex]
Donc
[tex]\sin(\dfrac{\pi}{8})=\sqrt{\dfrac{2-\sqrt{2}}{4}}=\dfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\\\\\\\boxed{\sin(\dfrac{\pi}{8})=\dfrac{\sqrt{2-\sqrt{2}}}{2}}[/tex]
b) Nous savons que : cos(a - b) = cos(a)cos(b) + sin(a)sin(b)
et sin(a - b) = sin(a)cos(b) - sin(b)cos(a).
Donc
[tex]\cos(\dfrac{3\pi}{8})=\cos(\dfrac{\pi}{2}-\dfrac{\pi}{8})\\\\\cos(\dfrac{3\pi}{8})=\cos(\dfrac{\pi}{2})\cos(\dfrac{\pi}{8})+\sin(\dfrac{\pi}{2})\sin(\dfrac{\pi}{8})\\\\\cos(\dfrac{3\pi}{8})=0\times\cos(\dfrac{\pi}{8})+1\times\sin(\dfrac{\pi}{8})\\\\\boxed{\cos(\dfrac{3\pi}{8})=\sin(\dfrac{\pi}{8})}[/tex]
et
[tex]\sin(\dfrac{3\pi}{8})=\sin(\dfrac{\pi}{2}-\dfrac{\pi}{8})\\\\\sin(\dfrac{3\pi}{8})=\sin(\dfrac{\pi}{2})\cos(\dfrac{\pi}{8})-\sin(\dfrac{\pi}{8})\cos(\dfrac{\pi}{2})\\\\\sin(\dfrac{3\pi}{8})=1\times\cos(\dfrac{\pi}{8})-\sin(\dfrac{\pi}{8})\times0\\\\\boxed{\sin(\dfrac{3\pi}{8})=\cos(\dfrac{\pi}{8})}[/tex]
[tex]c)\ f(\dfrac{\pi}{8})=\cos(\dfrac{3\pi}{8})\sin^3(\dfrac{\pi}{8})=\sin(\dfrac{\pi}{8})\sin^3(\dfrac{\pi}{8})=\sin^4(\dfrac{\pi}{8})\\\\\\=(\dfrac{\sqrt{2-\sqrt{2}}}{2})^4=\dfrac{(2-\sqrt{2})^2}{2^4}=\dfrac{4-4\sqrt{2}+2}{16}=\dfrac{6-4\sqrt{2}}{16}=\dfrac{3-2\sqrt{2}}{8}\\\\\Longrightarrow\boxed{f(\dfrac{\pi}{8})=\dfrac{3-2\sqrt{2}}{8}}[/tex]
et
[tex]f(\dfrac{3\pi}{8})=\cos(\dfrac{9\pi}{8})\sin^3(\dfrac{3\pi}{8})=-\cos(\dfrac{\pi}{8})\sin^3(\dfrac{3\pi}{8})\\\\\\=-\cos(\dfrac{\pi}{8})\cos^3(\dfrac{\pi}{8})=-\cos^4(\dfrac{\pi}{8})\\\\\\=-(\dfrac{\sqrt{2+\sqrt{2}}}{2})^4=-\dfrac{(2+\sqrt{2})^2}{2^4}=-\dfrac{4+4\sqrt{2}+2}{16}\\\\=-\dfrac{6+4\sqrt{2}}{16}=\dfrac{-3-2\sqrt{2}}{8}\\\\\Longrightarrow\boxed{f(\dfrac{3\pi}{8})=\dfrac{-3-2\sqrt{2}}{8}}[/tex]
2) a) La fonction f est 2π-périodique car pour tout x réel, f(x+2π) = f(x).
Nous pouvons alors restreindre l'étude de f à l'intervalle [-π ; π].
La fonction f est impaire car pour tout x réel, f(-x) = -f(x).
Le graphique représentant la fonction f est symétrique par rapport à l'origine (0;0) du repère.
Nous pouvons donc restreindre l'étude de f à l'intervalle [0 ; π].
Le graphique représentant la fonction f est symétrique par rapport au point (π/2 ; 0) car pour tout x réel, f(π/2 + x) + f(π/2 - x) = 0
Par conséquent, nous pouvons donc restreindre l'étude de f à l'intervalle [0 ; π/2].
[tex]b)\ f'(x)=[\cos(3x)\sin^3(x)]'=[\cos(3x)]'\times\sin^3(x)+\cos(3x)\times[\sin^3(x)]'\\\\=-3\sin(3x)\times\sin^3(x)+\cos(3x)\times[3\cos(x)\sin^2(x)]\\\\=3\sin^2(x)\times[-\sin(3x)\sin(x)+\cos(3x)\cos(x)]\\\\=3\sin^2(x)\times[\cos(3x)\cos(x)-\sin(3x)\sin(x)]\\\\=3\sin^2(x)\times\cos(3x+x)\\\\=3\sin^2(x)\times\cos(4x)\\\\\Longrightarrow\boxed{f'(x)=3\cos(4x)\sin^2(x)}[/tex]
c) Signe de f '(x) et variations de f sur l'intervalle [0 ; π/2]
Racines de f '(x)
[tex]3\cos(4x)\sin^2(x)=0\\\\\cos(4x)=0\ \ ou\ \ \sin(x)=0\\\\4x=\dfrac{\pi}{2}+k\pi\ \ ou\ \ x=k\pi\\\\Dans\ [0;\dfrac{\pi}{2}],\ \ x=\dfrac{\pi}{8}\ \ ou\ \ \ x=\dfrac{3\pi}{8}\ ou\ \ x=0\\\\\\\begin{array}{|c|ccccccc|} x&0&&\frac{\pi}{8}&&\frac{3\pi}{8}&&\frac{\pi}{2}\\3\cos(4x)&+&+&0&-&0&+&+\\\sin^2(x)&0&+&+&+&+&+&+\\f'(x)&0&+&0&-&0&+&+\\f'(x)&0&\nearrow&\frac{3-2\sqrt{2}}{8}&\searrow&\frac{-3-2\sqrt{2}}{8}&\nearrow&0\\\end{array}[/tex]
