Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Given $\sin x+\sin y+\sin z=0$ (1) and $\cos x+\cos y+\cos z=0$ (2). From (1), $\sin x+\sin y=-\sin z$. From (2), $\cos x+\cos y=-\cos z$. Squaring and adding these two equations:$$(\sin x+\sin y{)}^2+(\cos x+\cos y{)}^2=(-\sin z{)}^2+(-\cos z{)}^2$$ $$({\sin }^{2}x+{\cos }^{2}x)+({\sin }^{2}y+{\cos }^{2}y)+2(\sin x\sin y+\cos x\cos y)={\sin }^{2}z+{\cos }^{2}z$$ $$1+1+2\cos (x-y)=1⟹2\cos (x-y)=-1⟹\cos (x-y)=-\frac{1}{2}$$ Similarly, $\cos (y-z)=-\frac{1}{2}$ and $\cos (z-x)=-\frac{1}{2}$. This implies $x,y,z$ are angles of the form $\theta ,\theta +2\pi /3,\theta +4\pi /3$. We need to find ${\sin }^{2}x+{\sin }^{2}y+{\sin }^{2}z$. Using ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$:$${\sin }^{2}x+{\sin }^{2}y+{\sin }^{2}z=\frac{1-\cos 2x}{2}+\frac{1-\cos 2y}{2}+\frac{1-\cos 2z}{2}=\frac{3}{2}-\frac{1}{2}(\cos 2x+\cos 2y+\cos 2z)$$ For angles of the form $\theta ,\theta +2\pi /3,\theta +4\pi /3$, it is a known identity that $\cos 2\theta +\cos (2\theta +4\pi /3)+\cos (2\theta +8\pi /3)=0$. Therefore, $\cos 2x+\cos 2y+\cos 2z=0$. Substituting this back:$${\sin }^{2}x+{\sin }^{2}y+{\sin }^{2}z=\frac{3}{2}-\frac{1}{2}(0)=\frac{3}{2}$$