How do I find exact value of cos(2A) if cos A = sqrt(11)/6? And A is in Quadrant I?

Method 1 :

cos(2A)
= 2 cos²A - 1
= 2 × (√11/6)² - 1
= -14/36
= -7/18

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Method 2 :

Since A is in Quadrant I, sinA > 0
sin²A + cos²A = 1
sin²A + (√11/6)² = 1
sin²A + (11/36) = 1
sin²A = 1 - (11/36)
sin²A = 25/36
sinA = 5/6

cos(2A)
= cos(A + A)
= cos(A)cos(A) - sin(A)sin(A)
= (√11/6)² - (5/6)
= (11/36) - (25/36)
= -14/36
= -7/18

Do you know this identity?

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: b = a

cos(a + a) = cos(a).cos(a) - sin(a).sin(a)

cos(2a) = cos²(a) - sin²(a) → recall: cos²(a) + sin²(a) = 1 → sin²(a) = 1 - cos²(a)

cos(2a) = cos²(a) - [1 - cos²(a)]

cos(2a) = cos²(a) - 1 + cos²(a)

cos(2a) = 2.cos²(a) - 1 → given: cos(a) = (√11)/6

cos(2a) = 2.[(√11)/6]² - 1

cos(2a) = 2.[11/36] - 1

cos(2a) = (22/36) - (36/36)

cos(2a) = - 14/36

cos(2a) = - 7/18