Solveeit Logo
Login

Question

Question: Let A and B denote statements \( A:\cos \alpha + \cos \beta + \cos \gamma = 0 \\\ B:\sin \...

Let A and B denote statements
A:cos⁡α+cos⁡β+cos⁡γ=0 B:sin⁡α+sin⁡β+sin⁡γ=0  A:\cos \alpha + \cos \beta + \cos \gamma = 0 \\\ B:\sin \alpha + \sin \beta + \sin \gamma = 0 \\\
If cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−3/2\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

Explanation

Solution

In this question simplify the equation and add sin⁡2α+sin⁡2β+sin⁡2γ+cos⁡2α+cos⁡2β+cos⁡2γ{\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 (a+b+c)2=a2+b2+c2+2ab+2ac+bc{\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⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−3/2\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - 3/2 and we have to show that A:cos⁡α+cos⁡β+cos⁡γ=0 B:sin⁡α+sin⁡β+sin⁡γ=0 \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⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−3/2\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - 3/2we get
2[cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)]+3=02\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⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b
Therefore, 2[cos⁡βcos⁡γ+sin⁡βsin⁡γ+cos⁡γcos⁡α+sin⁡γsin⁡α+cos⁡αcos⁡β+sin⁡αsin⁡β]+3=02\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, forA:\cos \alpha + \cos \beta + \cos \gamma = 0andandB:\sin \alpha + \sin \beta + \sin \gamma = 0$ to be true is only possible
When cos⁡α+cos⁡β+cos⁡γ=0\cos \alpha + \cos \beta + \cos \gamma = 0 and sin⁡α+sin⁡β+sin⁡γ=0\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→Yf:X \to Y examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.