Question

If $sin\theta +2sin\varphi +3\sin \Psi =0$ and $cos\theta +2cos\varphi +3cos\Psi =0$; then evaluate
(a) $cos3\theta +8cos3\varphi +27cos3\Psi$
(b) $sin(\varphi +\Psi )+2sin(\Psi +\theta )+3sin(\theta +\Psi )$.

Answer 1

Put $a=cos\theta +isin\theta ;b=cos\varphi +isin\varphi ;c=cos\Psi +isin\Psi$
$\therefore a+2b+3c=(cos\theta +2cos\varphi +3cos\Psi )+i(sin\theta +2sin\varphi +3sin\Psi )$
$=0+i.0$ (given)
$\therefore a+2b+3c=0$
$\therefore a^3+8b^3+27c^3=18abc$
$\Rightarrow (cos\theta +isin\theta {)}^3+8(cos\varphi +isin\varphi {)}^3+27(cos\Psi +isin\Psi {)}^3$
$=18(cos\theta +isin\theta )(cos\varphi +isin\varphi )(cos\Psi +isin\Psi )$
$\Rightarrow (cos3\theta +isin3\theta )+8(cos3\varphi +isin3\varphi )+27(cos3\Psi +isin3\Psi )$
$=18\left\{cos\right(\theta +\varphi +\Psi )+isin(\theta +\varphi +\Psi \left)\right\}$
Equate the real parts, we get
$cos3\theta +8cos3\varphi +27cos3\Psi =18cos(\theta +\varphi +\Psi )$
Again $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=0$ (using hypothesis)
$\Rightarrow bc+2ca+3ab=0$
$\Rightarrow (cos\varphi +isin\varphi )(cos\Psi +isin\Psi )+2(cos\Psi +isin\Psi )(cos\theta +isin\theta )+3(cos\theta +isin\theta )(cos\varphi +isin\varphi )=0$
$\Rightarrow cos(\varphi +\Psi )+isin(\varphi +\Psi )+2cos(\Psi +\theta )+2isin(\Psi +\theta )+3cos(\theta +\varphi )+3isin(\theta -\varphi )=0$
Equate the imaginary parts, we get
$sin(\varphi +\Psi )+2sin(\Psi +\theta )+3sin(\theta +\varphi )=0$