A) 0
B) \[\cos (\alpha +\beta +\gamma )\]
C) \[3\cos (\alpha +\beta +\gamma )\]
D) \[3\sin (\alpha +\beta +\gamma )\]
Correct Answer: C
Solution :
\[\cos \alpha +\cos \beta +\cos \gamma =0\] and \[\sin \alpha +\sin \beta +\sin \gamma =0\] Let \[a=\cos \alpha +i\sin \alpha \,;\,b=\cos \beta +i\sin \beta \] and \[c=\cos \gamma +i\sin \gamma .\] Therefore \[a+b+c=(\cos \alpha +\cos \beta +\cos \gamma )\]\[+i\,(\sin \alpha +\sin \beta +\sin \gamma )\] \[=0+i0=0\] If \[a+b+c=0,\] then \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\] or \[{{(\cos \alpha +i\,\sin a)}^{3}}+{{(\cos \beta +i\sin \beta )}^{3}}+{{(\cos \gamma +i\sin \gamma )}^{3}}\]\[=3(\cos \alpha +i\sin \alpha )\,(\cos \beta +i\sin \beta )\,(\cos \gamma +i\sin \gamma )\] \[\Rightarrow (\cos 3\alpha +i\sin 3\alpha )+(\cos 3\beta +i\sin 3\beta )+(\cos 3\gamma +i\sin 3\gamma )\]\[=3[\cos (\alpha +\beta +\gamma )+i\sin (\alpha +\beta +\gamma )]\] or \[\cos 3\alpha +\cos 3\beta +\cos 3\gamma =3\cos (\alpha +\beta +\gamma ).\]
You need to login to perform this action.
You will be redirected in
3 sec
