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Question: Prove that: \[{\cos ^2}A + {\cos ^2}B - 2\cos A\cos B\cos \left( {A + B} \right) = {\sin ^2}\left(...

Prove that:
cos⁡2A+cos⁡2B−2cos⁡Acos⁡Bcos⁡(A+B)=sin⁡2(A+B){\cos ^2}A + {\cos ^2}B - 2\cos A\cos B\cos \left( {A + B} \right) = {\sin ^2}\left( {A + B} \right)

Explanation

Solution

In order to solve this question, use the angle sum formula on the left side of the equality containing the cos⁡(A+B)\cos \left( {A + B} \right)part. Then apply some operations and get it to an equation which can be considered as the common ground. Then, do the same for the right-hand side of the equality – apply the angle sum formula, then apply some operations, and finally get it to the same point as the cosine part on the left-hand side and then equate them.

Formula Used:
cos(A+B)=cosAcosB−sinAsinBcos\left( {A + B} \right) = cosAcosB-sinAsinB…(i)
sin(A+B)=sinAcosB+sinBcosAsin\left( {A + B} \right) = sinAcosB + sinBcosA…(ii)

Complete step by step solution:
First let us consider the left-hand side of the question – cos2A+cos2B−2cosAcosBcos(A+B)co{s^2}A + co{s^2}B-2cosAcosBcos\left( {A + B} \right)
Now, cos2A+cos2B−2cosAcosBcos(A+B)co{s^2}A + co{s^2}B-2cosAcosBcos\left( {A + B} \right)
=cos2A+cos2B−2cosAcosB(cosAcosB−sinAsinB)co{s^2}A + co{s^2}B-2cosAcosB\left( {cosAcosB - sinAsinB} \right) (using eq (i))
=cos⁡2A+cos⁡2B−2cos⁡2Acos⁡2B+2cos⁡Acos⁡Bsin⁡Asin⁡B{\cos ^2}A + {\cos ^2}B-2{\cos ^2}A{\cos ^2}B + 2\cos A\cos B\sin A\sin B …(iii)

Now, let us consider the right-hand side of the question – sin2(A+B)si{n^2}\left( {A + B} \right)
Now, sin2(A+B)si{n^2}\left( {A + B} \right)
= (sin(A+B))2{\left( {sin\left( {A + B} \right)} \right)^2}
Using the formula given above in (ii) we have:
= (sinAcosB+cosAsinB)2{\left( {sinAcosB + cosAsinB} \right)^2}
After opening up the square of the expression, we have:
= sin2Acos2B+cos2Asin2B+2sinAsinBcosAcosBsi{n^2}Aco{s^2}B + co{s^2}Asi{n^2}B + 2sinAsinBcosAcosB
Applying the formula –sin2A+cos2Asi{n^2}A + co{s^2}A=11, we get:
= (1−cos2A)cos2B+cos2A(1−cos2B)+2sinAsinBcosAcosB\left( {1 - co{s^2}A} \right)co{s^2}B + co{s^2}A\left( {1 - co{s^2}B} \right) + 2sinAsinBcosAcosB
= cos2B−cos2Acos2B+cos2A−cos2Acos2B+2sin⁡Asin⁡Bcos⁡Acos⁡Bco{s^2}B - co{s^2}Aco{s^2}B + co{s^2}A-co{s^2}Aco{s^2}B + 2\sin A \sin B \cos A \cos B
Rearranging and solving the equation we get:
=cos2A+cos2B−2cos2Acos2B+2sinAsinBcosAcosBco{s^2}A + co{s^2}B - 2co{s^2}Aco{s^2}B + 2sinAsinBcosAcosB …(iv)
On comparing equations (iii) and (iv), we get:
cos⁡2A+cos⁡2B−2cos⁡Acos⁡Bcos⁡(A+B){\cos ^2}A + {\cos ^2}B - 2\cos A\cos B\cos \left( {A + B} \right)=cos2A+cos2B−2cos2Acos2B+2sinAsinBcosAcosBco{s^2}A + co{s^2}B - 2co{s^2}Aco{s^2}B + 2sinAsinBcosAcosB
and also,
sin⁡2(A+B){\sin ^2}\left( {A + B} \right)=cos2A+cos2B−2cos2Acos2B+2sinAsinBcosAcosBco{s^2}A + co{s^2}B - 2co{s^2}Aco{s^2}B + 2sinAsinBcosAcosB
So, we can say that – L.H.S. = R.H.S.
Hence, proved.

Note:
In questions like these, if one finds it difficult to solve the exact answer, it is best advisable to first find a common ground for where both the equations could reach. That is, first solve the left-hand side up till the determined common ground. Then solve the right-hand side till the determined common ground and then since both of the sides of the equality get to the same point, i.e., they are equal, it can be said that the question is solved.