Trigonometry problem ( Restricted variables)

Question

I want to prove that if $x+y = z$ , then

$$\cos^2 x + \cos^2 y + \cos^2 z - 2\cos(x)\cos(y)\cos(z) = 1.$$

I have tried this question by putting z = x+y in the equation , but still there are a lot of terms left at the end ...

Answer 1

$$\cos^2 x + \cos^2 y + \cos^2 z - 2\cos(x)\cos(y)\cos(z) = $$ $$=\frac32+\frac{\cos 2x+\cos 2y+\cos 2z}{2}-2\cos(x)\cos(y)\cos(z)=$$ $$=\frac32+\frac{\cos 2x+\cos 2y}{2}+\frac{\cos 2z}{2}-2\cos(x)\cos(y)\cos(z)=$$ $$=\frac32+\cos (x+y)\cos(x-y)+\frac{2\cos^2 z-1}2-2\cos(x)\cos(y)\cos(z)=$$ $$=1+\cos z\left(\cos(x-y)+\cos(x+y)-2\cos(x)\cos(y)\right)=$$ $$=1+\cos z \left(2\cos x \cos y-2 \cos x \cos y\right)=1$$

Answer 2

\begin{align} & {{\cos }^{2}}z={{\cos }^{2}}(x+y)={{\left( \cos x\cos y-\sin x\sin y \right)}^{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,={{\cos }^{2}}x{{\cos }^{2}}y-\sin 2x\sin 2y+{{\sin }^{2}}x{{\sin }^{2}}y \\ \end{align} and \begin{align} & 2\cos x\cos y\cos z=2\cos x\cos y\cos (x+y) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\cos x\cos y(\cos x\cos y-\sin x\sin y) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2{{\cos }^{2}}x{{\cos }^{2}}y-\sin 2x\sin 2y \\ \end{align} we have [\begin{align} & I={{\cos }^{2}}x+{{\cos }^{2}}y+{{\cos }^{2}}z-2\cos x\cos y\cos z \\ & \,\,\,={{\cos }^{2}}x+{{\cos }^{2}}y+\underline{{{\cos }^{2}}x{{\cos }^{2}}y}-\sin 2x\sin 2y+{{\sin }^{2}}x{{\sin }^{2}}y\underline{-2{{\cos }^{2}}x{{\cos }^{2}}y}+\sin 2x\sin 2y \\ & \,\,\,={{\cos }^{2}}x+{{\cos }^{2}}y+{{\sin }^{2}}x{{\sin }^{2}}y-{{\cos }^{2}}x{{\cos }^{2}}y \\ & \,\,\,={{\cos }^{2}}x-{{\cos }^{2}}x{{\cos }^{2}}y+{{\cos }^{2}}y+{{\sin }^{2}}x{{\sin }^{2}}y \\ & \,\,\,={{\cos }^{2}}x(1-{{\cos }^{2}}y)+{{\cos }^{2}}y+{{\sin }^{2}}x{{\sin }^{2}}y \\ & \,\,\,={{\cos }^{2}}x{{\sin }^{2}}y+{{\cos }^{2}}y+{{\sin }^{2}}x{{\sin }^{2}}y \\ & \,\,\,=({{\cos }^{2}}x{{\sin }^{2}}y+{{\sin }^{2}}x{{\sin }^{2}}y)+{{\cos }^{2}}y \\ & \,\,\,={{\sin }^{2}}y\underbrace{({{\cos }^{2}}x+{{\sin }^{2}}x)}_{1}+{{\cos }^{2}}y \\ & \,\,\,={{\sin }^{2}}y+{{\cos }^{2}}y \\ & \,\,\,=\,1 \\ \end{align}

Answer 3

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$F=\cos^2x+\cos^2y+\cos^2z=1+\cos(x+y)\cos(x-y)+\cos z\cdot\cos z$$

If $\cos(x+y)=\cos z\iff x+y=2m\pi\pm z$ where $m$ is any integer

$$F=1+\cos z\cos(x-y)+\cos z\cos(x+y)=1+\cos z[\cos(x-y)+\cos(x+y)]=?$$