Question

If $\sum\limits_{A}{\cos A}=0=\sum\limits_{A}{\sin A}$ , then the value of ${{x}^{\sin (2A-B-C)}}{{x}^{\sin (2B-C-A)}}{{x}^{\sin (2C-A-B)}}$ should be equal to?
(a)1
(b)0
(c)3
(d)-3

Answer 1

Hint: In this question, we are given that the sum of the cosine and sine of the angles is equal to 0. Therefore, we can form 3 complex numbers whose argument will be equal to the given angles. Then, we can find out the difference between the arguments of the complex numbers which will be equal to the difference in the given angles, and as the exponents of x are given in terms of the different angles, we can put the values of the difference in angles to obtain the required answer.

Complete step-by-step answer:
In this question, the given angles are A, B and C and the relation between the angles is given to be
$\sum\limits_{A}{\cos A}=0=\sum\limits_{A}{\sin A}.........................(1.1)$
Now, if we form three complex numbers, ${{z}_{1}},{{z}_{2}}\text{ and }{{z}_{3}}$ with the given angles as their argument, we can write them as
$\begin{aligned} & {{z}_{1}}=\cos A+i\sin A \\\ & {{z}_{2}}=\cos B+i\sin B \\\ & {{z}_{3}}=\cos C+i\sin C.................................(1.2) \\\ \end{aligned}$
Taking the sum of the above complex numbers, we get
$\begin{aligned} & {{z}_{1}}+{{z}_{2}}+{{z}_{3}}=\cos A+i\sin A+\cos B+i\sin B+\cos C+i\sin C \\\ & =\left( \cos A+\cos B+\cos C \right)+i\left( \sin A+\sin B+\sin C \right) \\\ & =0+0=0...........................(1.3) \\\ \end{aligned}$
Where in the last line we have used equation (1.1) for the sum of the cosines and sines of the angles. Thus, from (1.3), we can write
${{z}_{1}}+{{z}_{2}}=-{{z}_{3}}..........................(1.4)$

Now, the formula for the magnitude of a complex number z=a+ib is given by
$\left| z \right|=\left| a+ib \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}.....................(1.5)$
Therefore, from (1.2) and using the fact that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ for any angle $\theta$ , we obtain
$\begin{aligned} & \left| {{z}_{1}} \right|=\sqrt{{{\cos }^{2}}A+{{\sin }^{2}}A}=\sqrt{1}=1 \\\ & \left| {{z}_{2}} \right|=\sqrt{{{\cos }^{2}}B+{{\sin }^{2}}B}=\sqrt{1}=1 \\\ & \left| {{z}_{3}} \right|=\sqrt{{{\cos }^{2}}C+{{\sin }^{2}}C}=\sqrt{1}=1...................(1.6) \\\ \end{aligned}$
Also, the magnitude of the difference of two complex numbers ${{z}_{1}}$ and ${{z}_{2}}$ is given by
${{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\cos \theta ........................(1.7)$
Using the values from (1.4) and (1.6) in (1.7), we obtain
$\begin{aligned} & {{\left| {{z}_{1}}+{{z}_{2}} \right|}^{2}}={{\left| {{z}_{3}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}+2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\cos \theta \\\ & \Rightarrow 1=1+1+2\times 1\times 1\cos \theta \\\ & \Rightarrow \cos \theta =\dfrac{-1}{2}........................(1.7) \\\ \end{aligned}$
However, we know that $\cos \left( \dfrac{2\pi }{3} \right)=\dfrac{-1}{2}..................(1.8)$
We can use the formula
$\begin{aligned} & \cos \left( \alpha \right)=\cos \left( \beta \right) \\\ & \Rightarrow \alpha =2n\pi \pm \beta \\\ \end{aligned}$
where n is an integer. However, as we are interested only in the difference of the two angles, we can take the angle within 0 and $2\pi$ and as we are interested about the relative orientation of the angles, we can take the coordinate axes such that the difference of their angles is positive. Therefore, equating the angles in (1.7) and (1.8) to obtain
$\theta =\dfrac{2\pi }{3}..................(1.9)$
However, we could also have written (1.4) as ${{z}_{1}}+{{z}_{3}}=-{{z}_{2}}$ or ${{z}_{2}}+{{z}_{3}}=-{{z}_{1}}$ and then followed the same procedure to obtain that the angle between the complex numbers ${{z}_{3}}$ and ${{z}_{2}}$, i.e. the difference of their arguments as $B-C=\dfrac{2\pi }{3}$
$\begin{aligned} & A-B=\dfrac{2\pi }{3} \\\ & B-C=\dfrac{2\pi }{3} \\\ & \Rightarrow A-C=A-B+B-C=\dfrac{2\pi }{3}+\dfrac{2\pi }{3}=\dfrac{4\pi }{3}..................(1.10) \\\ \end{aligned}$
As shown in the figure below
Therefore, we can now use these values in the expression given in the question to obtain

& {{x}^{\sin (2A-B-C)}}{{x}^{\sin (2B-C-A)}}{{x}^{\sin (2C-A-B)}}={{x}^{\sin (A-B+A-C)}}{{x}^{\sin (B-C+B-A)}}{{x}^{\sin (C-A+C-B)}} \\\ & ={{x}^{\sin \left( \dfrac{2\pi }{3}+\dfrac{4\pi }{3} \right)}}{{x}^{\sin \left( \dfrac{2\pi }{3}-\dfrac{2\pi }{3} \right)}}{{x}^{\sin \left( \dfrac{-4\pi }{3}+\dfrac{-2\pi }{3} \right)}} \\\ & ={{x}^{\sin \left( \dfrac{6\pi }{3} \right)}}{{x}^{\sin \left( 0 \right)}}{{x}^{\sin \left( \dfrac{-6\pi }{3} \right)}}={{x}^{\sin \left( 2\pi \right)}}{{x}^{\sin \left( 0 \right)}}{{x}^{\sin \left( -2\pi \right)}} \\\ & ={{x}^{0}}{{x}^{0}}{{x}^{0}}=1\times 1\times 1=1 \\\ \end{aligned}$$ Thus, we obtain the answer to the given question as 1 which matches option (a) and hence (a) is the correct answer to this question. Note: We should note that in equation (1.1), $\sum\limits_{A}{\cos A}=0=\sum\limits_{A}{\sin A}$ means that $\cos A+\cos B+\cos C=0=\sin A+\sin B+\sin C$ and the symbol A in the summation symbol just indicates that the variable A is being summed over. We could also have written the same expression using any other variable say P by $\sum\limits_{P}{\cos P}=0=\sum\limits_{P}{\sin P}$ and one should not get confused in the variable A used in the summation term and the angle A used in the given expression ${{x}^{\sin (2A-B-C)}}{{x}^{\sin (2B-C-A)}}{{x}^{\sin (2C-A-B)}}$ as they are different from each other.