Question

Let A and B denote statements
$
  A:\cos \alpha + \cos \beta + \cos \gamma = 0 \\
  B:\sin \alpha + \sin \beta + \sin \gamma = 0 \\
$
If $\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - 3/2$, then
(A). A is true and B is false
(B). A is false and B is true
(C). Both A and B are true
(D). Both A and B are false

Answer 1

Hint: In this question simplify the equation and add ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma + {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma $and also remember to use the algebraic formula ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2ac + bc$, using this information will help you to approach the solution of the question.

Complete step-by-step answer:
According to the given information we have equation $\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - 3/2$ and we have to show that $\begin{gathered}
  A:\cos \alpha + \cos \beta + \cos \gamma = 0 \\
  B:\sin \alpha + \sin \beta + \sin \gamma = 0 \\
\end{gathered} $
Simplifying the given equation $\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - 3/2$we get
$2\left[ {\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right)} \right] + 3 = 0$
As we know that according to the trigonometric identity $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$
Therefore, $2\left[ {\cos \beta \cos \gamma + \sin \beta \sin \gamma + \cos \gamma \cos \alpha + \sin \gamma \sin \alpha + \cos \alpha \cos \beta + \sin \alpha \sin \beta } \right] + 3 = 0$
$ \Rightarrow $\[2\cos \beta \cos \gamma + 2\sin \beta \sin \gamma + 2\cos \gamma \cos \alpha + 2\sin \gamma \sin \alpha + 2\cos \alpha \cos \beta + 2\sin \alpha \sin \beta + 3 = 0\]
Now adding ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma + {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma $ in the above equation we get
\[2\cos \beta \cos \gamma + 2\sin \beta \sin \gamma + 2\cos \gamma \cos \alpha + 2\sin \gamma \sin \alpha + 2\cos \alpha \cos \beta + 2\sin \alpha \sin \beta + {\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma + {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 0\]
As we know that ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2ac + bc$
Therefore, \[{\left( {\cos \alpha + \cos \beta + \cos \gamma } \right)^2} + {\left( {\sin \alpha + \sin \beta + \sin \gamma } \right)^2} = 0\]
So, for $A:\cos \alpha + \cos \beta + \cos \gamma = 0$ and $B:\sin \alpha + \sin \beta + \sin \gamma = 0$ to be true is only possible
When \[\cos \alpha + \cos \beta + \cos \gamma = 0\] and \[\sin \alpha + \sin \beta + \sin \gamma = 0\]
Therefore, both A and B are true
Hence, option C is the correct option.

Note: In the above solution we used the trigonometric identities which are the expressions which involve trigonometric functions where the term “function” can be explained as relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by $f:X \to Y$ examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.