If $ \cos x +2 \cos y+3 \cos z=0 , \sin x+2 \sin y+3 \sin z=0$ and $x+y+z=\pi$. Find $\sin 3x+8 \sin 3y+27 \sin 3z$

Question

Problem : If $ \cos x +2 \cos y+3 \cos z=0 , \sin x+2 \sin y+3 \sin z=0$ and $x+y+z=\pi$. Find $\sin 3x+8 \sin 3y+27 \sin 3z$

Solution: Adding $ \cos x +2 \cos y+3 \cos z=0$ and $\sin x+2 \sin y+3 \sin z=0$,we get

$ (\cos x+\sin x) +2(\cos y+\sin y)+3(\cos z+\sin z) =0$

Am I doing right ?

Answer 1

Here's another hint: $a^3 + b^3 + c^3 - 3abc = (a+b+c) (a^2 + b^2 +c^2 - ab-bc-ca)$.

We have $e^{ix} + 2e^{iy} + 3e^{iz} = 0 $.

Using the above identity, we get

$e^{i3x} + 8 e^{i3y} + 27e^{i3z} = 6e^{i(x+y+z)} = -6 $

Now compare real and imaginary parts.

Answer 2

Hint: use complex exponentials.