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Question: How do you prove \(\sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B\)?...

How do you prove sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B?

Explanation

Solution

In order to prove this we can use the formula of sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin \left( {A + B} \right) = \sin A\cos B + \sin B\cos A
And sin⁡(A−B)=sin⁡Acos⁡B−sin⁡Bcos⁡A\sin \left( {A - B} \right) = \sin A\cos B - \sin B\cos A so we can use this formula here and again use the general formula sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1 to prove the above property.

Complete step by step solution:
Here we are given to prove that sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B −−−−(1) - - - - (1)
So we need to formula of sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin \left( {A + B} \right) = \sin A\cos B + \sin B\cos A
And sin⁡(A−B)=sin⁡Acos⁡B−sin⁡Bcos⁡A\sin \left( {A - B} \right) = \sin A\cos B - \sin B\cos A
On substituting these formula in the given equation (1) and we will get:
sin⁡(A+B)sin⁡(A−B) (sin⁡Acos⁡B+sin⁡Bcos⁡A)(sin⁡Acos⁡B−sin⁡Bcos⁡A)−−−−−(2)  \sin (A + B)\sin (A - B) \\\ \left( {\sin A\cos B + \sin B\cos A} \right)\left( {\sin A\cos B - \sin B\cos A} \right) - - - - - (2) \\\
Now we have got the above equation of the form (a+b)(a−b)\left( {a + b} \right)\left( {a - b} \right)
So we can sue the formula (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
So using this formula in the equation (2) we will get the above equation as:
(sin⁡Acos⁡B+sin⁡Bcos⁡A)(sin⁡Acos⁡B−sin⁡Bcos⁡A)\left( {\sin A\cos B + \sin B\cos A} \right)\left( {\sin A\cos B - \sin B\cos A} \right)
sin⁡2Acos⁡2B−sin⁡2Bcos⁡2A{\sin ^2}A{\cos ^2}B - {\sin ^2}B{\cos ^2}A −−−(3) - - - (3)
Now the only thing which we need to think now is which value we have to substitute in order to get the above proof. Here in the above proof we can see that in the right hand side of the proof we need to get only the terms that are having the trigonometric function sin⁡\sin and not cos⁡\cos
So we need to use the formula where we can convert the trigonometric function cos⁡\cos into sin⁡\sin so that all the cos⁡\cos functions are removed from the equation (3).
Now we have the formula according to which we know that sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1
So form the above formula we will get:
cos⁡2A=1−sin⁡2A{\cos ^2}A = 1 - {\sin ^2}A
Also we can write:
cos⁡2B=1−sin⁡2B{\cos ^2}B = 1 - {\sin ^2}B
Now we can substitute the above two values of cos⁡2A and cos⁡2B{\cos ^2}A{\text{ and }}{\cos ^2}B in equation (3) and we will get:
sin⁡2Acos⁡2B−sin⁡2Bcos⁡2A{\sin ^2}A{\cos ^2}B - {\sin ^2}B{\cos ^2}A
sin⁡2A(1−sin⁡2B)−sin⁡2B(1−sin⁡2A){\sin ^2}A(1 - {\sin ^2}B) - {\sin ^2}B(1 - {\sin ^2}A)
Simplifying it further we will get:
sin⁡2A−sin⁡2Asin⁡2B−sin⁡2B+sin⁡2Asin⁡2B{\sin ^2}A - {\sin ^2}A{\sin ^2}B - {\sin ^2}B + {\sin ^2}A{\sin ^2}B
sin⁡2A−sin⁡2B{\sin ^2}A - {\sin ^2}B =RHS = {\text{RHS}}

Hence we have proved that sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B

Note:
Here in these types of problems where we need to prove the properties regarding the trigonometric functions, we must be aware and remembering all the general formula of the trigonometric functions like:
sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1
1+tan⁡2A=sec⁡2A 1+cot⁡2A=csc⁡2A  1 + {\tan ^2}A = {\sec ^2}A \\\ 1 + {\cot ^2}A = {\csc ^2}A \\\
Similarly more formulas are also there which the student must remember to solve such [problems of proof.