If sin theta + 2sin phi + 3sin psi = 0 and co | vedclass

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(B) Let $z_1 = \cos 	heta + i \sin 	heta$,$z_2 = \cos \phi + i \sin \phi$,and $z_3 = \cos \psi + i \sin \psi$. Given equations are $\cos 	heta + 2\

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$18\cos (	heta + \phi + \psi )$

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(B) Let $z_1 = \cos 	heta + i \sin 	heta$,$z_2 = \cos \phi + i \sin \phi$,and $z_3 = \cos \psi + i \sin \psi$. Given equations are $\cos 	heta + 2\cos \phi + 3\cos \psi = 0$ and $\sin 	heta + 2\sin \phi + 3\sin \psi = 0$. Combining these,we get $z_1 + 2z_2 + 3z_3 = 0$. Using the algebraic identity $x^3 + y^3 + z^3 = 3xyz$ if $x + y + z = 0$,where $x = z_1$,$y = 2z_2$,and $z = 3z_3$: $z_1^3 + (2z_2)^3 + (3z_3)^3 = 3(z_1)(2z_2)(3z_3) = 18 z_1 z_2 z_3$. By De Moivre's Theorem,$z_1^3 = \cos 3	heta + i \sin 3	heta$,$(2z_2)^3 = 8(\cos 3\phi + i \sin 3\phi)$,and $(3z_3)^3 = 27(\cos 3\psi + i \sin 3\psi)$. Thus,$(\cos 3	heta + 8\cos 3\phi + 27\cos 3\psi) + i(\sin 3	heta + 8\sin 3\phi + 27\sin 3\psi) = 18(\cos(	heta + \phi + \psi) + i \sin(	heta + \phi + \psi))$. Equating the real parts,we get $\cos 3	heta + 8\cos 3\phi + 27\cos 3\psi = 18\cos(	heta + \phi + \psi)$.

If $\sin \theta + 2\sin \phi + 3\sin \psi = 0$ and $\cos \theta + 2\cos \phi + 3\cos \psi = 0$,then the value of $\cos 3\theta + 8\cos 3\phi + 27\cos 3\psi = $

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