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

Prove that:
(cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2=4sin⁡2(x−y2){(\cos x - \cos y)^2} + {(\sin x - \sin y)^2} = 4{\sin ^2}\left( {\dfrac{{x - y}}{2}} \right)

Explanation

Solution

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⁡2θ=2cos⁡2θ1 - \cos 2\theta = 2{\cos ^2}\theta
cos⁡Acos⁡B+sin⁡Asin⁡B=cos⁡(A−B)\cos A\cos B + \sin A\sin B = \cos (A - B)
cos⁡2θ+sin⁡2θ=1{\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=4sin⁡2(x−y2){(\cos x - \cos y)^2} + {(\sin x - \sin y)^2} = 4{\sin ^2}\left( {\dfrac{{x - y}}{2}} \right)
Taking the left-hand side
⇒(cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2\Rightarrow {(\cos x - \cos y)^2} + {(\sin x - \sin y)^2}
Using (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab,
⇒cos⁡2x+cos⁡2y−2cos⁡xcos⁡y+sin⁡2x+sin⁡2y−2sin⁡xsin⁡y\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,
⇒cos⁡2x+sin⁡2x+cos⁡2y+sin⁡2x−2cos⁡xcos⁡y−2sin⁡xsin⁡y\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 cos2θ+sin⁡2θ=1co{s^2}\theta + {\sin ^2}\theta = 1, we get,
⇒1+1−2cos⁡xcos⁡y−2sin⁡xsin⁡y\Rightarrow 1 + 1 - 2\cos x\cos y - 2\sin x\sin y
On simplification we get,
⇒2−2cos⁡xcos⁡y−2sin⁡xsin⁡y\Rightarrow 2 - 2\cos x\cos y - 2\sin x\sin y
From the last two terms,
⇒2−2(cos⁡xcos⁡y+sin⁡xsin⁡y)\Rightarrow 2 - 2(\cos x\cos y + \sin x\sin y)
Using cos⁡Acos⁡B+sin⁡Asin⁡B=cos⁡(A−B)\cos A\cos B + \sin A\sin B = \cos (A - B), we get,
⇒2−2cos⁡(x−y)\Rightarrow 2 - 2\cos (x - y)
Taking 2 common,
⇒2[1−cos⁡(x−y)]\Rightarrow 2[1 - \cos (x - y)]
Now, using formula 1−cos⁡2θ=2cos⁡2θ1 - \cos 2\theta = 2{\cos ^2}\theta , we get,
⇒2[2cos⁡2(x−y2)]\Rightarrow 2\left[ {2{{\cos }^2}\left( {\dfrac{{x - y}}{2}} \right)} \right]
Now, simplifying the brackets
⇒4cos⁡2(x−y2) \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
⇒4cos⁡2(x−y2)\Rightarrow 4{\cos ^2}\left( {\dfrac{{x - y}}{2}} \right)
Using the formula,2cos⁡2θ=1−cos⁡2θ2{\cos ^2}\theta = 1 - \cos 2\theta
⇒2[1−cos⁡(x−y)]\Rightarrow 2[1 - \cos (x - y)]
Now, using cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos (A - B) = \cos A\cos B + \sin A\sin B
⇒2[1−(cos⁡xcos⁡y+sin⁡xsin⁡y)]\Rightarrow 2[1 - (\cos x\cos y + \sin x\sin y)]
On simplifying the brackets
⇒2−2(cos⁡xcos⁡y+sin⁡xsin⁡y)\Rightarrow 2 - 2(\cos x\cos y + \sin x\sin y)
⇒2−2cos⁡xcos⁡y−2sin⁡xsin⁡y\Rightarrow 2 - 2\cos x\cos y - 2\sin x\sin y
⇒1+1−2cos⁡xcos⁡y−2sin⁡xsin⁡y\Rightarrow 1 + 1 - 2\cos x\cos y - 2\sin x\sin y
Now, we know that cos2θ+sin⁡2θ=1co{s^2}\theta + {\sin ^2}\theta = 1
therefore substituting 1=cos2x+sin⁡2x1 = co{s^2}x + {\sin ^2}xand 1=cos2y+sin⁡2y1 = co{s^2}y + {\sin ^2}y
⇒cos⁡2x+sin⁡2x+cos⁡2y+sin⁡2y−2cos⁡xcos⁡y−2sin⁡xsin⁡y\Rightarrow {\cos ^2}x + {\sin ^2}x + {\cos ^2}y + {\sin ^2}y - 2\cos x\cos y - 2\sin x\sin y
⇒cos⁡2x+cos⁡2y−2cos⁡xcos⁡y+sin⁡2x+sin⁡2y−2sin⁡xsin⁡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 a2+b2−2ab=(a−b)2{a^2} + {b^2} - 2ab = {(a - b)^2}
⇒(cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)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