Question

If \(\cos(\theta -\phi) + \cos (\phi -\alpha) + \cos (\alpha-\theta) = \cfrac{-3}{2} \) then find the value of \(\sin \theta + \sin \phi + \sin \alpha \).

Answer 1

If \(\cos(\theta -\phi) + \cos (\phi -\alpha) + \cos (\alpha-\theta) = \cfrac{-3}{2} \) then find the value of \(\sin \theta + \sin \phi + \sin \alpha \).

^

Answer 2

i was thinking to multiply by 2 and then expand those cos (stuff)
perhaps that's the key to finding sin

Answer 3

too lazy to work out heheh

Answer 4

try what i just said^^
there will some fun things going xD

Answer 5

\(\large \begin{align} \color{black}{\cos(\theta -\phi) + \cos (\phi -\alpha) + \cos (\alpha-\theta) = \dfrac{-3}{2} \hspace{.33em}\\~\\
\implies \cos\theta\cos \phi +\sin\theta\sin \phi + \cos \phi \cos \alpha +\hspace{.33em}\\~\\
\sin\phi \sin \alpha+ \cos \alpha-\cos \theta = \dfrac{-3}{2} \hspace{.33em}\\~\\
\implies 2\cos\theta\cos \phi +2\sin\theta\sin \phi +2 \cos \phi \cos \alpha + \hspace{.33em}\\~\\
2\sin\phi \sin \alpha+ 2\cos \alpha \cos \theta+3 = 0 \hspace{.33em}\\~\\
\implies 2\cos\theta\cos \phi +2\sin\theta\sin \phi +2 \cos \phi \cos \alpha +2\sin\phi \sin \alpha+\hspace{.33em}\\~\\
2\cos \alpha \cos \theta+1+1+1 = 0 \hspace{.33em}\\~\\
\implies 2\cos\theta\cos \phi +2\sin\theta\sin \phi +2 \cos \phi \cos \alpha +2\sin\phi \sin \alpha+\hspace{.33em}\\~\\
2\cos \alpha \cos \theta+\cos^2\theta +\sin^2\theta+\cos^2\alpha+\sin^2\alpha +\cos^2\phi +\sin^2\phi= 0 \hspace{.33em}\\~\\
\implies (\cos \theta +\cos \alpha+\cos \phi )^2+(\sin \theta +\sin \alpha+\sin \phi )^2=0 \hspace{.33em}\\~\\
\implies \cos \theta +\cos \alpha+\cos \phi=0 \hspace{.33em}\\~\\
\implies \sin \theta +\sin \alpha+\sin \phi =0 \hspace{.33em}\\~\\
}\end{align}\)

Answer 6

very neat :)