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junior college 2 | H2 Maths
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Wee Teck
Wee Teck

junior college 2 chevron_right H2 Maths chevron_right Singapore

bii)

Date Posted: 4 years ago
Views: 250
J
J
4 years ago
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

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
Not sure if it’s something like this. Tried double angle formula the first time but I made a big mess. Approximations several times here are okay because theta is sufficiently small and its positive integral powers would be even smaller (negligible).
J
J
4 years ago
In your first part, the -2θ² in the 2nd last line became -θ² in the last line.

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 + ...
Wee Teck
Wee Teck
4 years ago
Thx!
Eric Nicholas K
Eric Nicholas K
4 years ago
Ouch
J
J
4 years ago
Anyway Wee Teck, here's the alternative working I posted on the main comment section (in case you missed it).

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
Wee Teck
Wee Teck
4 years ago
Yep i saw,thx!!