Cosine

The basic Double Angle Identity for Cosine states:

$cos(2\theta )=cos^{2}(\theta )-sin^{2}(\theta )$

The basic Pythagorean Identity states:

$sin^{2}(\theta )+cos^{2}(\theta )=1$

Lets replace Sine squared in the Double angle identity to try and match the identity in the problem

First, solve for Sine squared in the Pythagorean Identity.

$sin^{2}(\theta )=1-cos^{2}(\theta )$

Then replace Sine squared in the Double Angle Identity

$cos(2\theta )=cos^{2}(\theta )-(1-cos^{2}(\theta ))$

Simplify

$cos(2\theta ) = 2cos^{2}(\theta )-1$

You can see that $N=2$ and $M = 1$

$2+1 = 3$

The answer is $\boxed{3}.$

Using the cosine addition formula, $\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi$

Plugging in $\phi=\theta$ gives this to be $\cos2\theta=\cos^2\theta-\sin^2\theta$

Using a substitution for $\sin^2\theta$ with the Pythagorean Identity $\sin^2\theta+\cos^2\theta=1$ gives $\cos2\theta=2\cos^2\theta-1$ , so $M=1$ and $N=2$ . $M+N=\boxed{3}$ .

Double Angle Formula.♥

We know that ,Cos2 θ= 2cos sqθ-1 . . . . . . . (i) Given cos2θ=Ncos sqθ-M . . . . . . . . . . . . (ii) By comparing i &ii N=2 ,M=1 ,,, N+M=3

cos(2a) =2cos(a)-1 campair with the given equ. we get N=2 , M=1 N+M=2+1=3

since that double angle stated that \cos\2 theta = 2\cos^{2} - 1 where n=2 , m=1 so n+m must be equal 3

We know if sin^2 . A + cos^2 . A = 1 and cos 2A = cos^2 . A - sin^2 . A. So, cos 2A = N.cos^2 . A - M cos^2 . A - sin^2 . A. = N.cos^2 . A - M cos^2 . A - (1- cos^2 . A) = N.cos^2 . A - M 2cos^2 . A - 1 = N.cos^2 . A - M >>>>> N = 2 , M = 1 . Thus N + M = 1 + 2 = 3