Exercice 1 Calculer cos(7π12)cos\left(\dfrac{7\pi}{12}\right)cos(127π) et sin(π12)sin\left(\dfrac\pi{12}\right)sin(12π) Montrer que (∀x∈R) :cos(x)=cos(π3+x)+cos(π3−x)(\forall x\in\R)~: cos(x)=cos\left(\dfrac\pi3+x\right)+cos\left(\dfrac\pi3-x\right)(∀x∈R) :cos(x)=cos(3π+x)+cos(3π−x) Correction cos(7π12)=cos(π3+π4)=cos(π3)cos(π4)−sin(π3)sin(π4)=1222+3222=2+64\begin{align*} cos\left(\dfrac{7\pi}{12}\right)&=cos\left(\dfrac\pi3+\dfrac\pi4\right)\\ &=cos\left(\dfrac\pi3\right)cos\left(\dfrac\pi4\right)-sin\left(\dfrac\pi3\right)sin\left(\dfrac\pi4\right) \\ &=\dfrac12\dfrac{\sqrt{2}}2+\dfrac{\sqrt{3}}2\dfrac{\sqrt{2}}2 \\ &=\dfrac{\sqrt2+\sqrt6}{4} \end{align*}cos(127π)=cos(3π+4π)=cos(3π)cos(4π)−sin(3π)sin(4π)=2122+2322=42+6 sin(π12)=sin(π3−π4)=sin(π3)cos(π4)−cos(π3)sin(π4)=3212−3222=3−64\begin{align*} sin\left(\dfrac\pi{12}\right)&=sin\left(\dfrac\pi3-\dfrac\pi4\right)\\ &=sin\left(\dfrac\pi3\right)cos\left(\dfrac\pi4\right)-cos\left(\dfrac\pi3\right)sin\left(\dfrac\pi4\right) \\ &=\dfrac{\sqrt3}2\dfrac12-\dfrac{\sqrt{3}}2\dfrac{\sqrt{2}}2 \\ &=\dfrac{\sqrt3-\sqrt6}{4} \end{align*}sin(12π)=sin(3π−4π)=sin(3π)cos(4π)−cos(3π)sin(4π)=2321−2322=43−6 soit x∈Rx\in\Rx∈R cos(π3+x)=cos(π3)cosx−sin(π3)sinx=12cos(x)−32sin(x)\begin{align*} cos\left(\dfrac\pi3+x\right) &=cos\left(\dfrac\pi3\right)cos x-sin\left(\dfrac\pi3\right)sin x \\ &=\dfrac12cos(x)-\dfrac{\sqrt3}2sin(x) \end{align*}cos(3π+x)=cos(3π)cosx−sin(3π)sinx=21cos(x)−23sin(x) cos(π3−x)=cos(π3)cosx+sin(π3)sinx=12cos(x)+32sin(x)\begin{align*} cos\left(\dfrac\pi3-x\right) &=cos\left(\dfrac\pi3\right)cos x+sin\left(\dfrac\pi3\right)sin x \\ &=\dfrac12cos(x)+\dfrac{\sqrt3}2sin(x) \end{align*}cos(3π−x)=cos(3π)cosx+sin(3π)sinx=21cos(x)+23sin(x) Donc : cos(π3+x)+cos(π3−x)=cos(x)cos\left(\dfrac\pi3+x\right)+cos\left(\dfrac\pi3-x\right)=cos(x)cos(3π+x)+cos(3π−x)=cos(x)