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secondary 3 | A Maths
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Pls help to solve q5 d , thk
Date Posted: 2 years ago
Views: 384
(d)
sin(2y)/cos(2y) + 5 = 2• (1/cos(2y))²
sin(2y)•cos(2y) + 5•cos²(2y)
= 2•sin²(2y) + 2•cos²(2y)
2•sin²(2y) - sin(2y)•cos(2y) -3•cos²(2y) = 0
(2⋅sin(2y)-3⋅cos(2y))•(sin(2y) + cos(2y)) = 0
2⋅sin(2y)-3⋅cos(2y) = 0
tan(2y) = 1.5 ⇨
2y₁ = 56.31° → y₁ = 28.155
2y₂ = 236.31° → y₂ = 118.155
sin(2y) + cos(2y) = 0
tan(2y) = -1 ⇨
2y₃ = 135° → y₃ = 67.5°
2y₄ = 315° → y₄ = 157.5°
sin(2y)/cos(2y) + 5 = 2• (1/cos(2y))²
sin(2y)•cos(2y) + 5•cos²(2y)
= 2•sin²(2y) + 2•cos²(2y)
2•sin²(2y) - sin(2y)•cos(2y) -3•cos²(2y) = 0
(2⋅sin(2y)-3⋅cos(2y))•(sin(2y) + cos(2y)) = 0
2⋅sin(2y)-3⋅cos(2y) = 0
tan(2y) = 1.5 ⇨
2y₁ = 56.31° → y₁ = 28.155
2y₂ = 236.31° → y₂ = 118.155
sin(2y) + cos(2y) = 0
tan(2y) = -1 ⇨
2y₃ = 135° → y₃ = 67.5°
2y₄ = 315° → y₄ = 157.5°
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Thank a lot
Thank alot
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