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secondary 4 | A Maths
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Sonia
Sonia

secondary 4 chevron_right A Maths chevron_right Singapore

Hi!!! Could someone kindly help me with (iii) only ? Thanks so much :D

Date Posted: 4 years ago
Views: 372
J
J
4 years ago
cosA = 1 - 2 sin²(½A)
So sin²(½A) = (1 - cosA)/2

cosA = 2 cos²(½A) - 1
So cos²(½A) = (1 + cosA)/2


tan²(½A) = sin²(½A)/cos²(½A)

= (1 - cosA)/2 ÷ (1 + cosA)/2

= (1 - cosA)/(1 + cosA)

= (1 - cosA)/(1 + cosA) x (1 - cosA)/(1 - cosA)

= (1 - cosA)²/(1 - cos²A)

= (1 - cosA)²/sin²A



tan(½A)

= √[(1 - cosA)²/sin²A]

= (1 - cosA)/sinA

= (1 + 2/√5)/(1/√5)

= √5 + 2
Eric Nicholas K
Eric Nicholas K
4 years ago
Interesting approach
J
J
4 years ago
Actually this is just the standard derivation of the half angle formula

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syjiaxuan
Syjiaxuan's answer
290 answers (A Helpful Person)
Not sure if it’s correct
Eric Nicholas K
Eric Nicholas K
4 years ago
Yup! Of course it’s correct.
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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
Good evening Sonia! Here are my workings for this question. The O Levels are just two weeks away, I hope you are doing well in your revision!
Sonia
Sonia
4 years ago
Thank you! It’s nearing!!! Just 6 days left :/