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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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\

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 = $

Similar Questions

The product of all the roots of ${\left( \cos \frac{\pi }{3} + i\sin \frac{\pi }{3} \right)^{3/4}}$ is

If $\alpha, \beta$ are the roots of the equation $x^2 - 2x + 4 = 0$,then the value of $\alpha^n + \beta^n$ is

If $x + \frac{1}{x} = 2\cos \alpha$,then $x^n + \frac{1}{x^n} = $

If $z \in \mathbb{C}$ and $i z^3+4 z^2-z+4 i=0$,then a complex root of this equation having minimum magnitude is

If $\alpha, \beta$ are the roots of the equation $x^2-2x+4=0$ and for any $n \in N, \alpha^n+\beta^n=k \cos \frac{n \pi}{3}$, then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module