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Trigonometrical Equations
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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 = $
A
$\cos (3\theta + 3\phi + 3\psi )$
B
$18\cos (\theta + \phi + \psi )$
C
$6\cos (\theta + \phi + \psi )$
D
$36\cos (\theta + \phi + \psi )$
Solution
Let $a=\cos \theta+i \sin \theta, b=\cos \phi+i \sin \phi, c=\cos \psi+i \sin \psi$
$a+2 b+3 c=0+i .0$
$\mathrm{SO}$,
$a+2 b+3 c=0$
$a^{3}+8 b^{3}+27 c^{3}=3 \cdot(a \cdot 2 b \cdot 3 c)$
$=18 \mathrm{abc}$
$\mathrm{SO}$,
$\cos 3 \theta+8 . \cos 3 \phi+27 \cos \psi=18 \cos (\theta+\phi+\psi)$
Standard 11
Mathematics
