Réponses

2014-03-21T13:17:36+01:00
Bonjour,

1) M(\cos\theta;\sin\theta)\ et\ A(1;0)\\\\AM=\sqrt{(\cos\theta-1)^2+(\sin\theta-0)^2}\\\\AM=\sqrt{\cos^2\theta-2\cos\theta+1+\sin^2\theta}\\\\AM=\sqrt{(\cos^2\theta+\sin^2\theta)-2\cos\theta+1}\\\\AM=\sqrt{1-2\cos\theta+1}\\\\AM=\sqrt{2-2\cos\theta}\\\\AM=\sqrt{2(1-\cos\theta)}\\\\

AM=\sqrt{2\times2\sin^2\dfrac{\theta}{2}}\\\\AM=\sqrt{4\sin^2\dfrac{\theta}{2}}\\\\AM=2\sin\dfrac{\theta}{2}}

M(\cos\theta;\sin\theta)\ et\ B(0;1)

D'une part,

BM=\sqrt{(\cos\theta-0)^2+(\sin\theta-1)^2}\\\\BM=\sqrt{\cos^2\theta+\sin^2\theta-2\sin\theta+1}\\\\BM=\sqrt{(\cos^2\theta+\sin^2\theta)-2\sin\theta+1}\\\\BM=\sqrt{1-2\sin\theta+1}\\\\BM=\sqrt{2-2\sin\theta}\\\\BM=\sqrt{2(1-\sin\theta)}

BM=\sqrt{2(1-2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2})}\\\\BM=\sqrt{2(\sin^2\dfrac{\theta}{2}+
\cos^2\dfrac{\theta}{2}-2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2})}\\\\BM=\sqrt{2(\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2})^2}\\\\BM=\sqrt{2}(\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2})

D'autre part,

2\sin(\dfrac{\pi}{4}-\dfrac{\theta}{2})=2\times(\sin\dfrac{\pi}{4}\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2}\cos\dfrac{\pi}{4})\\\\2\sin(\dfrac{\pi}{4}-\dfrac{\theta}{2})=2\times(\dfrac{\sqrt{2}}{2}\cos\dfrac{\theta}{2}-\dfrac{\sqrt{2}}{2}\sin\dfrac{\theta}{2})\\\\2\sin(\dfrac{\pi}{4}-\dfrac{\theta}{2})=2\times\dfrac{\sqrt{2}}{2}(\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2})\\\\2\sin(\dfrac{\pi}{4}-\dfrac{\theta}{2})=\sqrt{2}(\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2})

Par conséquent,  BM=2\sin(\dfrac{\pi}{4}-\dfrac{\theta}{2})

2) AM+BM = 2\sin\dfrac{\theta}{2}}+2\sin(\dfrac{\pi}{4}-\dfrac{\theta}{2})\\\\AM+BM = 2[\sin\dfrac{\theta}{2}}+\sin(\dfrac{\pi}{4}-\dfrac{\theta}{2})]\\\\AM+BM = 2\times2\sin\dfrac{\dfrac{\theta}{2}+(\dfrac{\pi}{4}-\dfrac{\theta}{2})}{2}\cos\dfrac{\dfrac{\theta}{2}-(\dfrac{\pi}{4}-\dfrac{\theta}{2})}{2}\\\\AM+BM = 2\times2\sin\dfrac{\dfrac{\pi}{4}}{2}\cos\dfrac{\dfrac{\theta}{2}-\dfrac{\pi}{4}+\dfrac{\theta}{2}}{2}

AM+BM = 4\sin\dfrac{\dfrac{\pi}{4}}{2}\cos\dfrac{\theta-\dfrac{\pi}{4}}{2}\\\\AM+BM = 4\sin\dfrac{\pi}{8}\cos(\dfrac{\theta}{2}-\dfrac{\pi}{8}})

AM+BM sera maximale si  \cos(\dfrac{\theta}{2}-\dfrac{\pi}{8}})=1

\dfrac{\theta}{2}-\dfrac{\pi}{8}}=2k\pi\\\\\dfrac{\theta}{2}=\dfrac{\pi}{8}}+2k\pi\\\\\theta=2\times(\dfrac{\pi}{8}}+2k\pi)\\\\\theta=\dfrac{\pi}{4}}+4k\pi\\\\Or\ \ \theta\in]0;\dfrac{\pi}{2}[\\\\Donc\ \theta=\dfrac{\pi}{4}

AM + BM sera maximale si \theta=\dfrac{\pi}{4}