Entre que quelle que soit X à partir de R A . 1+sin(2x)= 2-(cos(x) - sin(x))² b . (1+cosx+sinx)²=2.... Pergunta de ideia deRihab11

May 2019 1 77 Report

Entre que quelle que soit X à partir de R
A . 1+sin(2x)= 2-(cos(x) - sin(x))²
b . (1+cosx+sinx)²=2(1+sinx )(1+cosx)

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Commentaires (2) Bonjour rihab11

$A)\ 2-(\cos(x) - \sin(x))^2=2-[\cos^2(x)+2\cos(x)\sin(x)+\sin^2(x)]\\\\2-(\cos(x) - \sin(x))^2=2-[(\cos^2(x)+\sin^2(x))+2\cos(x)\sin(x)]\\\\2-(\cos(x) - \sin(x))^2=2-[1+2\cos(x)\sin(x)\\\\2-(\cos(x) - \sin(x))^2=2-1-2\sin(x)\cos(x)\\\\2-(\cos(x) - \sin(x))^2=1-2\sin(x)\cos(x)\\\\\boxed{2-(\cos(x) - \sin(x))^2=1-\sin(2x)}\\\\\boxed{1-\sin(2x)=2-(\cos(x) - \sin(x))^2}$

$B)\ (1+\cos x+\sin x)^2=1+\cos^2x+\sin^2x\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2\cos x+2\sin x+2\cos x\sin x\\\\(1+\cos x+\sin x)^2=1+1+2\cos x+2\sin x+2\cos x\sin x\\\\(1+\cos x+\sin x)^2=2+2\cos x+2\sin x+2\cos x\sin x\\\\(1+\cos x+\sin x)^2=2(1+\cos x+\sin x+\cos x\sin x)\\\\(1+\cos x+\sin x)^2=2[(1+\cos x)+(\sin x+\cos x\sin x)]\\\\(1+\cos x+\sin x)^2=2[(1+\cos x)+\sin x(1+\cos x)]\\\\\boxed{(1+\cos x+\sin x)^2=2(1+\sin x)(1+\cos x)}$

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rihab11 merci beaucoup

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