Question

Let A and B denote the statements $A:\cos \alpha + \cos \beta + \cos \gamma = 0$, $B:\sin \alpha + \sin \beta + \sin \gamma = 0$. If $\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = \dfrac{{ - 3}}{2}$, then
$\left( 1 \right)$ A is correct and B is incorrect
$\left( 2 \right)$ A is incorrect and B is correct
$\left( 3 \right)$ both A and B are correct
$\left( 4 \right)$ both A and B are incorrect

Answer 1

We have to state that for the given condition which of the following statement/s is/are correct or incorrect. We solve this question using the concept of the values of the trigonometric functions and the concept of the various trigonometric formulas. We should also have the knowledge of the formula for the square of the sum of three numbers. First, we will simplify the given condition and then we will simplify the condition using the formula of square of sum of three numbers and the relation of the cosine and the sine function. Thus, using the properties of the squares of numbers we will compare the values and hence state that the given statements are correct or incorrect and hence get the required answer.

Complete step-by-step solution:
Given:
$\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = \dfrac{{ - 3}}{2}$
We can simplify the expression as:
$2\left[ {\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right)} \right] = - 3$
Further expanding and simplifying, we can write the expression as:
$2\cos \left( {\beta - \gamma } \right) + 2\cos \left( {\gamma - \alpha } \right) + 2\cos \left( {\alpha - \beta } \right) + 3 = 0$
$2\cos \left( {\beta - \gamma } \right) + 2\cos \left( {\gamma - \alpha } \right) + 2\cos \left( {\alpha - \beta } \right) + 1 + 1 + 1 = 0$
As, we know that the relation between cosine and sine is given as:
${\sin ^2}x + {\cos ^2}x = 1$
Using the formula, we can write the expression as:
$2\cos \left( {\beta - \gamma } \right) + 2\cos \left( {\gamma - \alpha } \right) + 2\cos \left( {\alpha - \beta } \right) + {\sin ^2}\alpha + {\cos ^2}\alpha + {\sin ^2}\beta + {\cos ^2}\beta + {\sin ^2}\gamma + {\cos ^2}\gamma = 0$
We also know that the formula for the square of sum of three numbers is given as:
${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$
Now, we are given the statements as:
$A:\cos \alpha + \cos \beta + \cos \gamma = 0$, $B:\sin \alpha + \sin \beta + \sin \gamma = 0$
Now, using the formula above we can expand the expression as:
${\left( {\cos \alpha + \cos \beta + \cos \gamma } \right)^2} = {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma + 2\cos \alpha \cos \beta + 2\cos \beta \cos \gamma + 2\cos \gamma \cos \alpha$
Also, we know that the formula for sum of two cosine terms is given as:
$2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$
Using the formula, we can write the expression as:
${\left( {\cos \alpha + \cos \beta + \cos \gamma } \right)^2} = {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma + \cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right) + \cos \left( {\beta + \gamma } \right) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma + \alpha } \right) + \cos \left( {\gamma - \alpha } \right)$
Similarly, for the sine statement
${\left( {\sin \alpha + \sin \beta + \sin \gamma } \right)^2} = {\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma + 2\sin \alpha \sin \beta + 2\sin \beta \sin \gamma + 2\sin \gamma \sin \alpha$
Also, we know that the formula for sum of two cosine terms is given as:
$2\sin A\sin B = \cos \left( {a - b} \right) - \cos \left( {a + b} \right)$
Using the formula, we can write the expression as:
${\left( {\sin \alpha + \sin \beta + \sin \gamma } \right)^2} = {\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma + \cos \left( {\alpha - \beta } \right) - \cos \left( {\alpha + \beta } \right) + \cos \left( {\beta - \gamma } \right) - \cos \left( {\beta + \gamma } \right) + \cos \left( {\gamma - \alpha } \right) - \cos \left( {\gamma + \alpha } \right)$
Now, adding the two statements and cancelling the terms, we get
${\left( {\sin \alpha + \sin \beta + \sin \gamma } \right)^2} + {\left( {\cos \alpha + \cos \beta + \cos \gamma } \right)^2} = {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma + {\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma + 2\cos \left( {\alpha - \beta } \right) + 2\cos \left( {\beta - \gamma } \right) + 2\cos \left( {\gamma - \alpha } \right)$
We have evaluated the value of the relation above as:
$2\cos \left( {\beta - \gamma } \right) + 2\cos \left( {\gamma - \alpha } \right) + 2\cos \left( {\alpha - \beta } \right) + {\sin ^2}\alpha + {\cos ^2}\alpha + {\sin ^2}\beta + {\cos ^2}\beta + {\sin ^2}\gamma + {\cos ^2}\gamma = 0$
Putting the value, we get the expression as:
${\left( {\sin \alpha + \sin \beta + \sin \gamma } \right)^2} + {\left( {\cos \alpha + \cos \beta + \cos \gamma } \right)^2} = 0$
As we know that the square of a number can never be negative. So, the above expression obtained is possible only if:
${\left( {\cos \alpha + \cos \beta + \cos \gamma } \right)^2} = 0$ or ${\left( {\sin \alpha + \sin \beta + \sin \gamma } \right)^2} = 0$
Taking the square root we get the expressions as:
$\cos \alpha + \cos \beta + \cos \gamma = 0$or $\sin \alpha + \sin \beta + \sin \gamma = 0$
Hence, both the given statements A and B are correct.
Thus, the correct option is $\left( 3 \right)$.

Note: These types of questions are highly conceptual and formula based. These questions have the maximum chances of getting calculation mistakes. We should take care of the sign while applying the formula of the difference of two cosine functions. A slight error would lead to a wrong solution.