ANAND KUMAR UPADHAYAY

How Cos (2A) = 2 Cos²A - 1 ? As we know that  ➡  Cos(2A) = Cos²A - Sin²A .......(1) Also Sin²A + Cos²A = 1  or Sin²A = 1 - Cos²A......[put in (1) eq.] ➡Cos (2A) = Cos²A - (1 - Cos²A)                 = Cos²A - 1 + Cos²A Hence,            Cos (2A) = 2 Cos²A - 1          or 1 + Cos(2A) = 2 Cos²A ➡ If angle is half i.e. θ = A then, Cos (A) = 2 Cos²(A/2) - 1       or 1 + Cos(A) = 2 Cos²(A/2)

How 1 - Cos2A = 2 sin²A? As we know that, Cos(2A) = Cos²A - Sin²A.........(1) ........👉{ Cos(2A) = Cos²A - Sin²A } also, Sin²A + Cos²A = 1       or Cos²A = 1- Sin²A......(2) ...............👉{ How Sin²θ + Cos²θ = 1?? } put eq. (2) into eq.(1) Cos(2A) = 1 - Sin²A - Sin²A Hence, Cos(2A) = 1 - Sin²A also, 1 - Cos2A = 2 sin²A  If angle(ϴ) is halved, then, CosA = 1 - Sin²(A/2) or  1 - CosA = 2 sin²(A/2)