J's answer to Rella's Junior College 1 H2 Maths Singapore question.
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∠ACB = π radians - ⅓π radians - θ radians = (⅔π - θ) radians
By the sine rule, AB / sin∠ACB = AC / sin∠ABC
x / sin (⅔π - θ) = y / sin ⅓π
y/x = sin ⅓π / sin (⅔π - θ)
y/x = √3/2 / (sin ⅔π cos θ - cos ⅔π sin θ)
[ Addition formula for sin(α - β) ]
y/x = √3/2 / (√3/2 cos θ - (-½) sin θ)
y/x = √3/2 / (√3/2 cos θ + ½ sin θ)
y/x = 1 / (cos θ + 1/√3 sin θ) (shown)
Second part in the comments.
By the sine rule, AB / sin∠ACB = AC / sin∠ABC
x / sin (⅔π - θ) = y / sin ⅓π
y/x = sin ⅓π / sin (⅔π - θ)
y/x = √3/2 / (sin ⅔π cos θ - cos ⅔π sin θ)
[ Addition formula for sin(α - β) ]
y/x = √3/2 / (√3/2 cos θ - (-½) sin θ)
y/x = √3/2 / (√3/2 cos θ + ½ sin θ)
y/x = 1 / (cos θ + 1/√3 sin θ) (shown)
Second part in the comments.
Date Posted:
3 years ago
y/x = 1 / (cos θ + 1/√3 sin θ)
If x is sufficiently small, then cos θ ≈ 1 - ½θ² and sin θ ≈ θ
y/x ≈ 1 / (1 - ½θ² + 1/√3 θ)
y/x ≈ (1 + θ/√3 - ½θ²)⁻¹
From here, use the Maclaurin expansion for (1 + x)ⁿ . Replace x with θ/√3 - ½θ²
y/x ≈ 1 + (-1)(θ/√3 - ½θ²) + (-1)(-2)/2! (θ/√3 - ½θ²)² + ...
(The subsequent terms are not needed since we need quadratic expression, so we only need up to θ²)
y/x ≈ 1 - θ/√3 + ½θ² + (θ/√3)² + ...
( No need to expand the (θ/√3 - ½θ²)² fully )
y/x ≈ 1 - θ/√3 + ½θ² + θ²/3 + ...
y/x ≈ 1 - θ/√3 + 5/6 θ² + ...
If x is sufficiently small, then cos θ ≈ 1 - ½θ² and sin θ ≈ θ
y/x ≈ 1 / (1 - ½θ² + 1/√3 θ)
y/x ≈ (1 + θ/√3 - ½θ²)⁻¹
From here, use the Maclaurin expansion for (1 + x)ⁿ . Replace x with θ/√3 - ½θ²
y/x ≈ 1 + (-1)(θ/√3 - ½θ²) + (-1)(-2)/2! (θ/√3 - ½θ²)² + ...
(The subsequent terms are not needed since we need quadratic expression, so we only need up to θ²)
y/x ≈ 1 - θ/√3 + ½θ² + (θ/√3)² + ...
( No need to expand the (θ/√3 - ½θ²)² fully )
y/x ≈ 1 - θ/√3 + ½θ² + θ²/3 + ...
y/x ≈ 1 - θ/√3 + 5/6 θ² + ...
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