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junior college 2 | H2 Maths
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bii)
= (cos2θ + sin2θ)/[(cos2θ - sin2θ)(cos2θ + sin2θ)]
(Rationalising the denominator)
= (cos2θ + sin2θ)/(cos²2θ - sin²2θ)
= (cos2θ + sin2θ) / cos4θ
Using the Maclaurin's series,
QR = [(1 - (2θ)²/2! + (2θ)⁴/4! - (2θ)^6 /6! +...) + (2θ - (2θ)³/3! + (2θ)^5 /5! - (2θ)⁴/4! + ... )] / (1 - (4θ)²/2! + (4θ)⁴/4! - (4θ)^6 /6! +...)
Since θ is sufficiently small, powers of x³ and higher can be neglected.
So,
QR ≈ (1 - (2θ)²/2! + 2θ) / (1 - (4θ)²/2!)
QR ≈ (1 - 2θ² + 2θ)/(1 - 8θ²)
QR ≈ (1 - 2θ² + 2θ)(1 + (-8θ²))-¹
≈ (1 - 2θ² + 2θ)(1 + 8θ² + ...)
(Use series again , this time for (1 + x)ⁿ )
≈ (1 + 8θ² - 2θ² + 2θ + ...)
≈ 1 + 2θ + 6θ²
a = 2, b = 6
See 1 Answer
Final answer should be 1 + 2θ + 6θ²
The method is correct though.
Software's expansion gives
1 + 2θ + 6θ² + 44/3 θ³ + 38 θ⁴ + (1444 θ^5)/15 + (1228 θ^6)/5 + (196888 θ^7)/315 + (167158 θ^8)/105 + (11492164 θ^9)/2835 + (48775364 θ^10)/4725 + (4098748888 θ^11)/155925 + ...
QR = 1/(cos2θ - sin2θ)
= (cos2θ + sin2θ)/[(cos2θ - sin2θ)(cos2θ + sin2θ)]
(Rationalising the denominator)
= (cos2θ + sin2θ)/(cos²2θ - sin²2θ)
= (cos2θ + sin2θ) / cos4θ
Using the Maclaurin's series,
QR = [(1 - (2θ)²/2! + (2θ)⁴/4! - (2θ)^6 /6! +...) + (2θ - (2θ)³/3! + (2θ)^5 /5! - (2θ)⁴/4! + ... )] / (1 - (4θ)²/2! + (4θ)⁴/4! - (4θ)^6 /6! +...)
Since θ is sufficiently small, powers of x³ and higher can be neglected.
So,
QR ≈ (1 - (2θ)²/2! + 2θ) / (1 - (4θ)²/2!)
QR ≈ (1 - 2θ² + 2θ)/(1 - 8θ²)
QR ≈ (1 - 2θ² + 2θ)(1 + (-8θ²))-¹
≈ (1 - 2θ² + 2θ)(1 + 8θ² + ...)
(Use series again , this time for (1 + x)ⁿ )
≈ (1 + 8θ² - 2θ² + 2θ + ...)
≈ 1 + 2θ + 6θ²
a = 2, b = 6