If x + y + z = 180^o, then cos 2x + cos 2y - | vedclass

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(B) Given: $x + y + z = 180^o \implies x + y = 180^o - z$.Consider the expression: $\cos 2x + \cos 2y - \cos 2z$.Using the formula $\cos C + \cos D =

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$1 - 4\sin x \sin y \cos z$

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(B) Given: $x + y + z = 180^o \implies x + y = 180^o - z$.Consider the expression: $\cos 2x + \cos 2y - \cos 2z$.Using the formula $\cos C + \cos D = 2\cos(\frac{C+D}{2})\cos(\frac{C-D}{2})$:$\cos 2x + \cos 2y = 2\cos(x+y)\cos(x-y)$.Using the identity $\cos 2z = 2\cos^2 z - 1$:Expression $= 2\cos(x+y)\cos(x-y) - (2\cos^2 z - 1)$.Since $x+y = 180^o - z$,then $\cos(x+y) = \cos(180^o - z) = -\cos z$.Substituting this into the expression:$= 2(-\cos z)\cos(x-y) - 2\cos^2 z + 1$$= 1 - 2\cos z [\cos(x-y) + \cos z]$.Since $z = 180^o - (x+y)$,then $\cos z = \cos(180^o - (x+y)) = -\cos(x+y)$.$= 1 - 2\cos z [\cos(x-y) - \cos(x+y)]$.Using the identity $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$:$= 1 - 2\cos z [2\sin x \sin y]$$= 1 - 4\sin x \sin y \cos z$.

If $x + y + z = 180^o,$ then $\cos 2x + \cos 2y - \cos 2z$ is equal to

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