Complex Numbers And Quadratic Equations question 218
Question: If $ \sin \alpha +\sin \beta +\sin \gamma =0= $ $ \cos \alpha +\cos \beta +\cos \gamma , $ then the value of $ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma $ is [RPET 1999]
Options:
A) 2/3
B) 3/2
C) 1/2
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
$ \cos \alpha +\cos \beta +\cos \gamma =0 $ ??(i) $ \sin \alpha +\sin \beta +\sin \gamma =0 $ ………(ii) Let $ a=\cos \alpha +i\sin \alpha , $ $ b=\cos \beta +i\sin \beta $ $ c=\cos \gamma +i\sin \gamma $
$ \Rightarrow $ $ a+b+c=0 $ [by (i) and (ii)] …..(iii) $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} $ = $ {{[\cos \alpha +i\sin \alpha ]}^{-1}}+{{[\cos \beta +i\sin \beta ]}^{-1}}+{{[\cos \gamma +i\sin \gamma ]}^{-1}} $ $ =\cos \alpha -i\sin \alpha $ $ +\cos \beta -i\sin \beta $ $ +\cos \gamma -i\sin \gamma $
$ \Rightarrow ab+bc+ca=0 $ ….(iv) [by (i) and (ii)] Squaring both sides of equation (iii), we get $ a^{2}+b^{2}+c^{2}+2ab+2bc+2ca=0 $ or $ a^{2}+b^{2}+c^{2}=0 $ [by (iv)]
$ \therefore {{[\cos \alpha +i\sin \alpha ]}^{2}}+{{[\cos \beta +i\sin \beta ]}^{2}}+{{[\cos \gamma +i\sin \gamma ]}^{2}}=0 $
$ \Rightarrow (\cos 2\alpha +\cos 2\beta +\cos 2\gamma ) $ $ +i(\sin 2\alpha +\sin 2\beta +\sin 2\gamma )=0 $ separation of real and imaginary part ,
$ \Rightarrow \cos 2\alpha +\cos 2\beta +\cos 2\gamma =0 $ …..(v) and $ \sin 2\alpha +\sin 2\beta +\sin 2\gamma =0 $ …..(vi)
$ \Rightarrow 1-2{{\sin }^{2}}\alpha +1-2{{\sin }^{2}}\beta +1-2{{\sin }^{2}}\gamma =0 $ [by eq. (v)]
$ \therefore {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma =\frac{3}{2} $ .
BETA
