Question

If cos x + cos y + cos z = 0 & sin x + sin y + sin z = 0 then find the value of $\cot \left( {\dfrac{{x + y}}{2}} \right)$
A) $\sin z$
B) $\cos z$
C) $\cot z$
D) $2\sin z$

Answer 1

Hint: Try to get 2 seperate values of $\cos z$ and $\sin z$ from the two equations already given, the values will be in terms of $\cos x,\cos y\& \sin x,\sin y$ then divide $\cos z$ by $\sin z$ to get the value of $cot z$.

Complete Step by Step Solution:
We are given that $\cos x + \cos y + \cos z = 0\& \sin x + \sin y + \sin z = 0$
So from $\cos x + \cos y + \cos z = 0$
We can get $\cos x + \cos y = - \cos z$
Now break it into $\cos x + \cos y$
So we will get it as

\cos x + \cos y = - \cos z\\\ 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2} = - \cos z...............................................(i) \end{array}$$ Again from the sine part we will get $$\begin{array}{l} \sin x + \sin y = - \sin z\\\ 2\sin \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2} = - \sin z.................................................(ii) \end{array}$$ Now if we divide equation (i) by equation (ii) We will get it as $$\begin{array}{l} \therefore \dfrac{{2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}}}{{2\sin \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}}} = \dfrac{{ - \cos z}}{{ - \sin z}}\\\ \Rightarrow \dfrac{{\cos \dfrac{{x + y}}{2}}}{{\sin \dfrac{{x + y}}{2}}} = \dfrac{{\cos z}}{{\sin z}}\\\ \Rightarrow \cot \left( {\dfrac{{x + y}}{2}} \right) = \cot z \end{array}$$ Therefore option C is the correct option here. Note: $$\cos (x + y) = 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}\& \sin (x + y) = 2\sin \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}$$ These formulas are often used so remember them. Also in the solution we are trying to find the value of $$\cot z$$ which is equal to $$\dfrac{{\cos z}}{{\sin z}}$$ students commonly divide with the multiplicative inverse i.e., $$\dfrac{{\sin z}}{{\cos z}}$$ which leads to an incorrect solution.