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An0nyme1
@An0nyme1
May 2019
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Bonjour,
J’ai besoin d’aide pour cet exercice de trigonométrie (exercice 17 page 161, collection Math’x, éditions Didier).
Merci d’avance.
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Verified answer
A(0) = (cos 0 + sin 0)² + (cos 0 - sin 0)² → (1+0)² + (1-0)² = 2
A(π/4)=(cos π/4 + sin π/4)² + (cos π/4 - sin π/4)² → (√2/2+√2/2)² +(√2/2-√2)
²= 2
A(π/4)=(cos π/3 + sin π/3)² + (cos π/3 - sin π/3)² → (1/2+√3/2)² +(1/2-√3/2)
²= 2
∀x , A(x) = (cos x + sin x)² + (cos x - sin x)² →
A(x) = cos²x + 2sinxcosx + sin²x + cos²x - 2sinxcosx + sin²x = 2sin²x+2cos²x
A(x) = 2(sin²x+cos²x) = 2(1) = 2
Donc ∀x , (cos x + sin x)² + (cos x - sin x)
² = 2
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Verified answer
A(0) = (cos 0 + sin 0)² + (cos 0 - sin 0)² → (1+0)² + (1-0)² = 2A(π/4)=(cos π/4 + sin π/4)² + (cos π/4 - sin π/4)² → (√2/2+√2/2)² +(√2/2-√2)²= 2
A(π/4)=(cos π/3 + sin π/3)² + (cos π/3 - sin π/3)² → (1/2+√3/2)² +(1/2-√3/2)²= 2
∀x , A(x) = (cos x + sin x)² + (cos x - sin x)² →
A(x) = cos²x + 2sinxcosx + sin²x + cos²x - 2sinxcosx + sin²x = 2sin²x+2cos²x
A(x) = 2(sin²x+cos²x) = 2(1) = 2
Donc ∀x , (cos x + sin x)² + (cos x - sin x)² = 2