Question

Prove that:
${(\cos x - \cos y)^2} + {(\sin x - \sin y)^2} = 4{\sin ^2}\left( {\dfrac{{x - y}}{2}} \right)$

Answer 1

Hint: For a question like this we approach the solution by simplifying any one side and proving it equal to the other side, here also we’ll simplify the left-hand side using some trigonometric formulas like
$1 - \cos 2\theta = 2{\cos ^2}\theta $
$\cos A\cos B + \sin A\sin B = \cos (A - B)$
${\cos ^2}\theta + {\sin ^2}\theta = 1$
We simplify in such a manner that it results in equivalent value to the other side expression.

Complete step by step Answer:

Given data:${(\cos x - \cos y)^2} + {(\sin x - \sin y)^2} = 4{\sin ^2}\left( {\dfrac{{x - y}}{2}} \right)$
Taking the left-hand side
$ \Rightarrow {(\cos x - \cos y)^2} + {(\sin x - \sin y)^2}$
Using ${(a - b)^2} = {a^2} + {b^2} - 2ab$,
$ \Rightarrow {\cos ^2}x + {\cos ^2}y - 2\cos x\cos y + {\sin ^2}x + {\sin ^2}y - 2\sin x\sin y$
On rearranging we get,
$ \Rightarrow {\cos ^2}x + {\sin ^2}x + {\cos ^2}y + {\sin ^2}x - 2\cos x\cos y - 2\sin x\sin y$
Now, using $co{s^2}\theta + {\sin ^2}\theta = 1$, we get,
$ \Rightarrow 1 + 1 - 2\cos x\cos y - 2\sin x\sin y$
On simplification we get,
$ \Rightarrow 2 - 2\cos x\cos y - 2\sin x\sin y$
From the last two terms,
$ \Rightarrow 2 - 2(\cos x\cos y + \sin x\sin y)$
Using $\cos A\cos B + \sin A\sin B = \cos (A - B)$, we get,
$ \Rightarrow 2 - 2\cos (x - y)$
Taking 2 common,
$ \Rightarrow 2[1 - \cos (x - y)]$
Now, using formula $1 - \cos 2\theta = 2{\cos ^2}\theta $, we get,
$ \Rightarrow 2\left[ {2{{\cos }^2}\left( {\dfrac{{x - y}}{2}} \right)} \right]$
Now, simplifying the brackets
$ \Rightarrow 4{\cos ^2}\left( {\dfrac{{x - y}}{2}} \right)$, which is equal to the right-hand side in the given equation
Since, Left-hand side=right-hand side
We have proved the given equation

Note: An alternative method for the solution of the given question can be
 This time we’ll simplify the term in the right-hand side and will prove it equal to the term in the left-hand side
$ \Rightarrow 4{\cos ^2}\left( {\dfrac{{x - y}}{2}} \right)$
Using the formula,$2{\cos ^2}\theta = 1 - \cos 2\theta $
$ \Rightarrow 2[1 - \cos (x - y)]$
Now, using $\cos (A - B) = \cos A\cos B + \sin A\sin B$
$ \Rightarrow 2[1 - (\cos x\cos y + \sin x\sin y)]$
On simplifying the brackets
$ \Rightarrow 2 - 2(\cos x\cos y + \sin x\sin y)$
$ \Rightarrow 2 - 2\cos x\cos y - 2\sin x\sin y$
$ \Rightarrow 1 + 1 - 2\cos x\cos y - 2\sin x\sin y$
Now, we know that $co{s^2}\theta + {\sin ^2}\theta = 1$
therefore substituting \[1 = co{s^2}x + {\sin ^2}x\]and $1 = co{s^2}y + {\sin ^2}y$
$ \Rightarrow {\cos ^2}x + {\sin ^2}x + {\cos ^2}y + {\sin ^2}y - 2\cos x\cos y - 2\sin x\sin y$
$ \Rightarrow {\cos ^2}x + {\cos ^2}y - 2\cos x\cos y + {\sin ^2}x + {\sin ^2}y - 2\sin x\sin y$
Now, using the formula ${a^2} + {b^2} - 2ab = {(a - b)^2}$
\[ \Rightarrow {\left( {\cos x - \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2}\]which is equal to the left-hand side in the given equation
Since, Left-hand side=right-hand side
We have proved the given equation