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Question: Prove the following equation: \({\left( {\cos a + \cos b} \right)^2} + {\left( {\sin a + \sin b} \ri...

Prove the following equation: (cos⁡a+cos⁡b)2+(sin⁡a+sin⁡b)2=4×cos⁡2(a−b2){\left( {\cos a + \cos b} \right)^2} + {\left( {\sin a + \sin b} \right)^2} = 4 \times {\cos ^2}\left( {\dfrac{{a - b}}{2}} \right)

Explanation

Solution

We will take each side and try to simplify them separately using different trigonometric identities. We should be familiar with different trigonometric identities like cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b\cos (a - b) = \cos a\cos b + \sin a\sin b , cos⁡2x=2cos⁡2x−1\cos 2x = 2{\cos ^2}x - 1 etc. The most important thing is how we can equate both sides of the equation.

Complete step by step answer:
We have to prove (cos⁡a+cos⁡b)2+(sin⁡a+sin⁡b)2=4×cos⁡2(a−b2){\left( {\cos a + \cos b} \right)^2} + {\left( {\sin a + \sin b} \right)^2} = 4 \times {\cos ^2}\left( {\dfrac{{a - b}}{2}} \right)
We will first take L.H.S. and simplify it then try to convert our R.H.S. equals to L.H.S.
Now we will take our L.H.S.
(cos⁡a+cos⁡b)2+(sin⁡a+sin⁡b)2{\left( {\cos a + \cos b} \right)^2} + {\left( {\sin a + \sin b} \right)^2}
We will expand the square term
(cos⁡2a+cos⁡2b+2cos⁡acos⁡b)+(sin⁡2a+sin⁡2b+2sin⁡asin⁡b)\left( {{{\cos }^2}a + {{\cos }^2}b + 2\cos a\cos b} \right) + \left( {{{\sin }^2}a + {{\sin }^2}b + 2\sin a\sin b} \right)
We know that sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 , so we have
2+2cos⁡acos⁡b+2sin⁡asin⁡b2 + 2\cos a\cos b + 2\sin a\sin b
We have taken 2 common from second and third term
2+2(cos⁡acos⁡b+sin⁡asin⁡b)2 + 2\left( {\cos a\cos b + \sin a\sin b} \right)
We know that cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b\cos (a - b) = \cos a\cos b + \sin a\sin b, we get
2+2cos⁡(a−b)2 + 2\cos \left( {a - b} \right)
We will take our R.H.S.
4×cos⁡2(a−b2)4 \times {\cos ^2}\left( {\dfrac{{a - b}}{2}} \right)
2×2cos⁡2(a−b2)2 \times 2{\cos ^2}\left( {\dfrac{{a - b}}{2}} \right)
We know that cos⁡2x=2cos⁡2x−1\cos 2x = 2{\cos ^2}x - 1 ,we get
2×(cos⁡(a−b)+1)2 \times \left( {\cos \left( {a - b} \right) + 1} \right)
We will remove the bracket
=2+2cos⁡(a−b)= 2 + 2\cos \left( {a - b} \right)
Which is equal to L.H.S.
Hence, we have proved that (cos⁡a+cos⁡b)2+(sin⁡a+sin⁡b)2=4×cos⁡2(a−b2){\left( {\cos a + \cos b} \right)^2} + {\left( {\sin a + \sin b} \right)^2} = 4 \times {\cos ^2}\left( {\dfrac{{a - b}}{2}} \right)

Note:
We should know what are different trigonometric ratios and formulas. Sometimes we can simplify the question but are not able to equate both parts. We can start this question from taking RHS as well and that will not affect the final answer.