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Question: Using addition and subtraction identity prove that \[\sin \left( {A + B} \right).\sin \left( {A - B}...

Using addition and subtraction identity prove that sin⁡(A+B).sin⁡(A−B)=sin⁡2A−sin⁡2B\sin \left( {A + B} \right).\sin \left( {A - B} \right) = {\sin ^2}A - {\sin ^2}B

Explanation

Solution

Hint : This is simply a problem related to sum and difference identities. We will start with the LHS of the question and with the help of identities will proceed towards the RHS. We will try to make the terms in terms of sin function because the RHS is in sin function only.
Formula used:
sin⁡(A+B)=sin⁡A.cos⁡B+cos⁡A.sin⁡B\sin \left( {A + B} \right) = \sin A.\cos B + \cos A.\sin B
sin⁡(A−B)=sin⁡A.cos⁡B−cos⁡A.sin⁡B\sin \left( {A - B} \right) = \sin A.\cos B - \cos A.\sin B

Complete step-by-step answer :
We can directly start with the LHS as,
sin⁡(A+B).sin⁡(A−B)\sin \left( {A + B} \right).\sin \left( {A - B} \right)
Now we will write the formulas expansion as,
=(sin⁡A.cos⁡B+cos⁡A.sin⁡B)(sin⁡A.cos⁡B−cos⁡A.sin⁡B)= \left( {\sin A.\cos B + \cos A.\sin B} \right)\left( {\sin A.\cos B - \cos A.\sin B} \right)
Now will multiply the terms in first bracket with those in the second bracket,
=sin⁡A.cos⁡B(sin⁡A.cos⁡B−cos⁡A.sin⁡B)+cos⁡A.sin⁡B(sin⁡A.cos⁡B−cos⁡A.sin⁡B)= \sin A.\cos B\left( {\sin A.\cos B - \cos A.\sin B} \right) + \cos A.\sin B\left( {\sin A.\cos B - \cos A.\sin B} \right)
On multiplying we get,
=sin⁡2A.cos⁡2B−sin⁡A.cos⁡B.cos⁡A.sin⁡B+cos⁡A.sin⁡B.sin⁡A.cos⁡B−cos⁡2A.sin⁡2B= {\sin ^2}A.{\cos ^2}B - \sin A.\cos B.\cos A.\sin B + \cos A.\sin B.\sin A.\cos B - {\cos ^2}A.{\sin ^2}B
On careful observation we come to the conclusion that the second and third term are opposites of each other. So on cancelling them we get,
=sin⁡2A.cos⁡2B−cos⁡2A.sin⁡2B= {\sin ^2}A.{\cos ^2}B - {\cos ^2}A.{\sin ^2}B
Now we can observe half of the RHS in the answer above only need to remove the cos function. So as we know,
cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta
Applying this to equation above,
=sin⁡2A(1−sin⁡2B)−(1−sin⁡2A)sin⁡2B= {\sin ^2}A\left( {1 - {{\sin }^2}B} \right) - \left( {1 - {{\sin }^2}A} \right){\sin ^2}B
On multiplying again,
=sin⁡2A−sin⁡2A.sin⁡2B−sin⁡2B+sin⁡2A.sin⁡2B= {\sin ^2}A - {\sin ^2}A.{\sin ^2}B - {\sin ^2}B + {\sin ^2}A.{\sin ^2}B
Cancelling the opposite terms,
=sin⁡2A−sin⁡2B= {\sin ^2}A - {\sin ^2}B
And this is ,
=RHS= RHS
Hence proved.

Note : Note that this is a very basic problem as the question itself helps us to lead towards the problem. Only the thing is when we multiply the brackets we should be careful about the terms and the signs. One wrong sign can collapse the proof and just wastes the time. Along with sum and difference formulas we should be familiar with factorization and defactorization formulae.