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secondary 4 | A Maths
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Hi!!! Could someone kindly help me with (iii) only ? Thanks so much :D
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
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