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

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(B) Given $x + y + z = 180^\circ$,so $x + y = 180^\circ - z$.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 $x + y + z = 180^\circ$,so $x + y = 180^\circ - z$.Consider the expression $\cos 2x + \cos 2y - \cos 2z$.Using the formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:$\cos 2x + \cos 2y = 2 \cos(x + y) \cos(x - y)$.Since $x + y = 180^\circ - z$,$\cos(x + y) = \cos(180^\circ - z) = -\cos z$.Using the identity $\cos 2z = 2 \cos^2 z - 1$:$\cos 2x + \cos 2y - \cos 2z = 2(-\cos z) \cos(x - y) - (2 \cos^2 z - 1)$$= 1 - 2 \cos z \cos(x - y) - 2 \cos^2 z$$= 1 - 2 \cos z (\cos(x - y) + \cos z)$.Since $z = 180^\circ - (x + y)$,$\cos z = \cos(180^\circ - (x + y)) = -\cos(x + y)$.$= 1 - 2 \cos z (\cos(x - y) - \cos(x + y))$.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 $x + y + z = 180^\circ$,then $\cos 2x + \cos 2y - \cos 2z$ is equal to

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