Question

 If expression $\cos x+\cos y=\dfrac{1}{3}$ and $\sin x+\sin y=\dfrac{1}{4}$, prove that $\tan \left( \dfrac{x+y}{2} \right)=\dfrac{3}{4}$.

Answer 1

Hint: To prove the equality, if we start form the given equalities, then we should know few of the trigonometric addition formulae which are, $\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$ and $\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$. By using these relations, we can prove the desired relation given in the question.
Complete step-by-step answer:
In this question, we have been asked to prove $\tan \left( \dfrac{x+y}{2} \right)=\dfrac{3}{4}$, given that if $\cos x+\cos y=\dfrac{1}{3}$ and $\sin x+\sin y=\dfrac{1}{4}$.
To prove the desired relation, we will first start with $\sin x+\sin y=\dfrac{1}{4}$. We know that $\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$. So, by applying it in the above, we can write it as, $\Rightarrow2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)=\dfrac{1}{4}\ldots \ldots \ldots \left( i \right)$
Now, we will consider $\cos x+\cos y=\dfrac{1}{3}$. We know that $\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$. So, we can apply it in the above expression and we can write it as,$2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)=\dfrac{1}{3}\ldots \ldots \ldots \left( ii \right)$
Now, we will divide equation (i) by equation (ii). Dividing the equations, we get, $\Rightarrow\dfrac{2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)}{2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)}=\dfrac{\dfrac{1}{4}}{\dfrac{1}{3}}$
Now, we can see that, in the left hand side, $2\cos \left( \dfrac{x-y}{2} \right)$ is common in both the numerator as well as the denominator. So, we can cancel these like terms. After cancelling of the like terms, we get, $\Rightarrow\dfrac{\sin \left( \dfrac{x+y}{2} \right)}{\cos \left( \dfrac{x+y}{2} \right)}=\dfrac{3}{4}$
Now, we also know that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $. So, by substituting it in the above equation, we get,$\tan \left( \dfrac{x+y}{2} \right)=\dfrac{3}{4}$, which was asked to be proven in the question.
Hence we have proved that, $\tan \left( \dfrac{x+y}{2} \right)=\dfrac{3}{4}$ if $\cos x+\cos y=\dfrac{1}{3}$ and $\sin x+\sin y=\dfrac{1}{4}$.

Note: To solve this question, there is a possibility that the approach of using the addition formulas like, $\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$ and $\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$, do not strike our mind. So, we have to think of approaching the solution in all possible directions to get the desired result.