Exercice 9 --- (id : 910)
Trigonométrie: Exercice 9
correction

Rappel

🔸cos2x+sin2x=1\cos^2x+\sin^2x=1
🔸tanx=sinxcosx\tan x=\dfrac{\sin x}{\cos x}

1 1+tan2x1+\tan^2x =1+(sinxcosx)2=1+\left({\dfrac{\sin x}{\cos x}}\right)^2 =1+sin2xcos2x=1+\dfrac{\sin^2x}{\cos^2x} =cos2xcos2x+sin2xcos2x=\dfrac{\cos^2x}{\cos^2x}+\dfrac{\sin^2x}{\cos^2x} =cos2x+sin2xcos2x=1cos2x=\dfrac{\cos^2x+\sin^2x}{\cos^2x}=\dfrac{1}{\cos^2x} =sin2x+cos2xsin2x=\dfrac{\sin^2x+\cos^2x}{\sin^2x}
2 1+1tan2x1+\dfrac{1}{\tan^2x} =1+1sin2xcos2x=1+\dfrac{1}{\dfrac{\sin^2x}{\cos^2x}} =1+cos2xsin2x=1+\dfrac{\cos^2x}{\sin^2x} =sin2xsin2x+cos2xsin2x=\dfrac{\sin^2x}{\sin^2x}+\dfrac{\cos^2x}{\sin^2x} =sin2x+cos2xsin2x=\dfrac{\sin^2x+\cos^2x}{\sin^2x} =1sin2x=\dfrac{1}{\sin^2x}
3 cos2xsin2x\cos^2x-\sin^2x =(1sin2x)sin2x=(1-\sin^2x)-\sin^2x =12sin2x=1-2\sin^2x
4 (cosx+sinx)2(\cos x+\sin x)^2 =cos2x+sin2x1+2cosxsinx=\underbrace{\cos^2 x+\sin^2 x}_{1}+2\cos x\sin x =1+2cosxsinx=1+2\cos x\sin x
5 (cosxsinx)2(\cos x-\sin x)^2 =cos2x+sin2x12cosxsinx=\underbrace{\cos^2 x+\sin^2 x}_{1}-2\cos x\sin x =12cosxsinx=1-2\cos x\sin x