Question
Let A and B denote the statements
$$\eqalign{
& {\bf{A}}\,:\cos \alpha + \cos \beta + \cos \gamma = 0 \cr
& {\bf{B}}\,:\sin \alpha + \sin \beta + \sin \gamma = 0 \cr} $$
If $$\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - \frac{3}{2},$$ then:
A.
A is false and B is true
B.
both A and B are true
C.
both A and B are false
D.
A is true and B is false
Answer :
both A and B are true
Solution :
We have
$$\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - \frac{3}{2}$$
$$\eqalign{
& \Rightarrow \,\,2\left[ {\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right)} \right] + 3 = 0 \cr
& \Rightarrow \,\,2\left[ {\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right)} \right] + {\sin ^2}\alpha + {\cos ^2}\alpha + {\sin ^2}\beta + {\cos ^2}\beta + {\sin ^2}\gamma + {\cos ^2}\gamma = 0 \cr
& \Rightarrow \,\,\left[ {{{\sin }^2}\alpha + {{\sin }^2}\beta + {{\sin }^2}\gamma + 2\sin \alpha \sin \beta + 2\sin \beta \sin \gamma + 2\sin \gamma \sin \alpha } \right] + \left[ {{{\cos }^2}\alpha + {{\cos }^2}\beta + {{\cos }^2}\gamma + 2\cos \alpha \cos \beta + 2\cos \beta \cos \gamma + 2\cos \gamma \cos \alpha } \right] = 0 \cr
& \Rightarrow \,\,{\left[ {\sin \alpha + \sin \beta + \sin \gamma } \right]^2} + {\left( {\cos \alpha + \cos \beta + \cos \gamma } \right)^2} = 0 \cr
& \Rightarrow \,\,\sin \alpha + \sin \beta + \sin \gamma = 0\,\,{\text{and }}\cos \alpha + \cos \beta + \cos \gamma = 0 \cr} $$
∴ A and B both are true.