Question

Using various trigonometric identities and properties, prove the following results.
${{\left( \cos x+\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}=4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$ .

Answer 1

Hint: We will use the following formula $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ and $\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ also, ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. Using all these trigonometric identities, along with some basic manipulations, we will solve this question and will try to show that LHS is equal to RHS.

Complete step-by-step answer:
It is given in the question that we have to prove ${{\left( \cos x+\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}=4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$.
We will solve the LHS first using some basic trigonometric formulas as follows$\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ and $\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$.
In LHS, we get,
$={{\left( \cos x+\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}}$
Using the above identities, we get
$={{\left\{ 2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right) \right\}}^{2}}+{{\left\{ 2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right) \right\}}^{2}}$
Opening the squares, we get,
$=4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right){{\cos }^{2}}\left( \dfrac{x-y}{2} \right)+4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right){{\sin }^{2}}\left( \dfrac{x-y}{2} \right)$
Now taking $4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$ common from both the terms in the LHS, we get,
$=4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)\left\{ {{\cos }^{2}}\left( \dfrac{x-y}{2} \right)+{{\sin }^{2}}\left( \dfrac{x-y}{2} \right) \right\}$
Now, we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$, using this relation in LHS, we get LHS reduced to
$=4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)\left\{ 1 \right\}$
$=4{{\cos }^{2}}\left( \dfrac{x+y}{2} \right)$
This is equal to the RHS of the statement we have to prove.
Therefore, \[LHS=RHS\].
Hence proved.

Note: Since there are a very large number of trigonometric identities, there is not a single solution to such proves. These results can be proven in many alternative steps and by using various other identities. Therefore, it is important to learn all the basic identities of the trigonometric identities and formulas to solve these kinds of problems. We can remember all the basic formulas. Not only the formulas and identities, but also memorize the condition during which they hold true. There are many questions, where only conditions decide the answers, and no calculations are required. Also, in this question, if we do calculations step by step, silly mistakes can be avoided.