Question

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

Answer 1

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

Complete step by step solution:
Here we are given to prove that $\sin (A + B)\sin (A - B) = {\sin ^2}A - {\sin ^2}B$$ - - - - (1)$
So we need to formula of $\sin \left( {A + B} \right) = \sin A\cos B + \sin B\cos A$
 And $\sin \left( {A - B} \right) = \sin A\cos B - \sin B\cos A$
On substituting these formula in the given equation (1) and we will get:
$
  \sin (A + B)\sin (A - B) \\
  \left( {\sin A\cos B + \sin B\cos A} \right)\left( {\sin A\cos B - \sin B\cos A} \right) - - - - - (2) \\
 $
Now we have got the above equation of the form $\left( {a + b} \right)\left( {a - b} \right)$
So we can sue the formula $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
So using this formula in the equation (2) we will get the above equation as:
$\left( {\sin A\cos B + \sin B\cos A} \right)\left( {\sin A\cos B - \sin B\cos A} \right)$
${\sin ^2}A{\cos ^2}B - {\sin ^2}B{\cos ^2}A$$ - - - (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 $ and not $\cos $
So we need to use the formula where we can convert the trigonometric function $\cos $ into $\sin $ so that all the $\cos $ functions are removed from the equation (3).
Now we have the formula according to which we know that ${\sin ^2}A + {\cos ^2}A = 1$
So form the above formula we will get:
${\cos ^2}A = 1 - {\sin ^2}A$
Also we can write:
${\cos ^2}B = 1 - {\sin ^2}B$
Now we can substitute the above two values of ${\cos ^2}A{\text{ and }}{\cos ^2}B$ in equation (3) and we will get:
${\sin ^2}A{\cos ^2}B - {\sin ^2}B{\cos ^2}A$
${\sin ^2}A(1 - {\sin ^2}B) - {\sin ^2}B(1 - {\sin ^2}A)$
Simplifying it further we will get:
${\sin ^2}A - {\sin ^2}A{\sin ^2}B - {\sin ^2}B + {\sin ^2}A{\sin ^2}B$
${\sin ^2}A - {\sin ^2}B$$ = {\text{RHS}}$

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

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:
${\sin ^2}A + {\cos ^2}A = 1$
$
  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.