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ayoubgamouss13
@ayoubgamouss13
January 2021
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S'il vous plaît aidez moi c'est un devoir maison
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Geijutsu
Verified answer
Bonsoir,
-------------------------------------------------------
Rappels de cours :
∀x∈ℝ, sin²(x)+cos²(x) = 1
∀x∈ℝ\{(π/2)+πℤ}, sec(x) = 1/(cos(x)), donc cos(x)sec(x) = 1
∀x∈ℝ\{πℤ}, csc(x) = 1/(sin(x)), donc sin(x)csc(x) = 1
∀x∈ℝ\{πℤ/2}, sec²(x)+csc²(x) = sec²(x)csc²(x)
-------------------------------------------------------
a) ∀x∈ℝ,
sin(x)(sin(x)+cos(x))-cos(x)(sin(x)-cos(x)) = sin²(x)+sin(x)cos(x)-sin(x)cos(x)+cos²(x) = sin²(x)+cos²(x) =
1
b) ∀x∈ℝ\{πℤ/2},
B = (cos(x)+sec(x))²+(sin(x)+csc(x))²-(sec(x)csc(x))² = cos²(x)+2+sec²(x)+sin²(x)+2+csc²(x)-sec²(x)csc²(x) = cos²(x)+sin²(x)+2+2+sec²(x)+csc²(x)-sec²(x)csc²(x) = 1+2+2+sec²(x)csc²(x)-sec²(x)csc²(x) = 5+sec²(x)csc²(x)-sec²(x)csc²(x) =
5
1 votes
Thanks 1
ayoubgamouss13
Merci beaucoup
Geijutsu
Pour la dernière question, x∈ℝ\{πℤ/2}
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ayoubgamouss13
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Calculer : tan 73 fois tan 17 - sin au carré 40 - sin au carré 50 S'il vous plaît aidez moiii
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ayoubgamouss13
January 2021 | 0 Respostas
ABC un triangle rectangle en A, tel que Ab=4 et Bc=8 et AC =4racine 3. H le point de [Bc] tel que Bh = 5 La perpendiculaire à (Bc) menée par H coupe (Ab) en K. Calculer Hk S'il vous plaît aidez moi
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Lista de comentários
Verified answer
Bonsoir,-------------------------------------------------------
Rappels de cours :
∀x∈ℝ, sin²(x)+cos²(x) = 1
∀x∈ℝ\{(π/2)+πℤ}, sec(x) = 1/(cos(x)), donc cos(x)sec(x) = 1
∀x∈ℝ\{πℤ}, csc(x) = 1/(sin(x)), donc sin(x)csc(x) = 1
∀x∈ℝ\{πℤ/2}, sec²(x)+csc²(x) = sec²(x)csc²(x)
-------------------------------------------------------
a) ∀x∈ℝ,
sin(x)(sin(x)+cos(x))-cos(x)(sin(x)-cos(x)) = sin²(x)+sin(x)cos(x)-sin(x)cos(x)+cos²(x) = sin²(x)+cos²(x) = 1
b) ∀x∈ℝ\{πℤ/2},
B = (cos(x)+sec(x))²+(sin(x)+csc(x))²-(sec(x)csc(x))² = cos²(x)+2+sec²(x)+sin²(x)+2+csc²(x)-sec²(x)csc²(x) = cos²(x)+sin²(x)+2+2+sec²(x)+csc²(x)-sec²(x)csc²(x) = 1+2+2+sec²(x)csc²(x)-sec²(x)csc²(x) = 5+sec²(x)csc²(x)-sec²(x)csc²(x) = 5