Trigonométrie : Exercice 9 2ème année secondaire

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24 exercices

Exercice 9 --- (id : 910)

Trigonométrie: Exercice 9

correction

Rappel

🔸 $\cos^2x+\sin^2x=1$
🔸 $\tan x=\dfrac{\sin x}{\cos x}$

1

1+\tan^2x

=1+\left({\dfrac{\sin x}{\cos x}}\right)^2

=1+\dfrac{\sin^2x}{\cos^2x}

=\dfrac{\cos^2x}{\cos^2x}+\dfrac{\sin^2x}{\cos^2x}

=\dfrac{\cos^2x+\sin^2x}{\cos^2x}=\dfrac{1}{\cos^2x}

=\dfrac{\sin^2x+\cos^2x}{\sin^2x}

2

1+\dfrac{1}{\tan^2x}

=1+\dfrac{1}{\dfrac{\sin^2x}{\cos^2x}}

=1+\dfrac{\cos^2x}{\sin^2x}

=\dfrac{\sin^2x}{\sin^2x}+\dfrac{\cos^2x}{\sin^2x}

=\dfrac{\sin^2x+\cos^2x}{\sin^2x}

=\dfrac{1}{\sin^2x}

3

\cos^2x-\sin^2x

=(1-\sin^2x)-\sin^2x

=1-2\sin^2x

4

(\cos x+\sin x)^2

=\underbrace{\cos^2 x+\sin^2 x}_{1}+2\cos x\sin x

=1+2\cos x\sin x

5

(\cos x-\sin x)^2

=\underbrace{\cos^2 x+\sin^2 x}_{1}-2\cos x\sin x

=1-2\cos x\sin x

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