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secondary 3 | A Maths
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Date Posted: 2 years ago
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(a)
sin2x = 2sinx•cosx
2•sin2x - 3•cosx = 0
4•sinx•cosx - 3•cosx = 0
cosx•(4•sinx - 3) = 0
⇨ cosx = 0
⇨ sinx = ¾
x₁ = 90°
x₂ = 270°
x₃ = 48.59°
x₄ = 131.41°
sin2x = 2sinx•cosx
2•sin2x - 3•cosx = 0
4•sinx•cosx - 3•cosx = 0
cosx•(4•sinx - 3) = 0
⇨ cosx = 0
⇨ sinx = ¾
x₁ = 90°
x₂ = 270°
x₃ = 48.59°
x₄ = 131.41°
(b)
cos2θ = 1 - 2•sin²θ
sinθ = a
2•(1 - 2a²) + 3a -2 = 0
4a² -3a = 0
a•(4a-3) = 0
⇨ a = sinθ = 0
⇨a = sinθ = ¾
θ₁ = 0°
θ₂ = 180°
θ₃ = 360°
θ₄ = 48.59°
θ₅ = 131.41°
cos2θ = 1 - 2•sin²θ
sinθ = a
2•(1 - 2a²) + 3a -2 = 0
4a² -3a = 0
a•(4a-3) = 0
⇨ a = sinθ = 0
⇨a = sinθ = ¾
θ₁ = 0°
θ₂ = 180°
θ₃ = 360°
θ₄ = 48.59°
θ₅ = 131.41°
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Why change 2sin2x to 4sin cos x
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