How do you prove $2 {cos}^{2} A - 1 = cos (2 A)$ $2 cos^2 A - 1 = cos(2A)$ ?

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Answer 1 · 2016-03-14T01:06:16

Well we know that for two angles $A, B$ $A,B$ it holds that

$cos (A + B) = cos A cos B - sin A \cdot sin B$ $cos(A+B)=cosAcosB-sinA*sinB$

hence for $A = B$ $A=B$ you get

$cos (2 A) = {cos}^{2} A - {sin}^{2} A$ $cos(2A)=cos^2A-sin^2A$

But ${sin}^{2} A = 1 - {cos}^{2} A$ $sin^2A=1-cos^2A$

hence

$cos (2 A) = {cos}^{2} A - (1 - {cos}^{2} A) = 2 {cos}^{2} A - 1$ $cos(2A)=cos^2A-(1-cos^2A)=2cos^2A-1$

How do you prove 2cos2A−1=cos(2A)2 cos^2 A - 1 = cos(2A)?