Question

Prove that Sin (A+B) × Sin (A-B) = Sin²A - Sin²B

Answer 1

Hey there !

Solution :

We know the formulas for all the values in the above question. They are :

Sin ( A + B ) = Sin A . Cos B + Cos A . Sin B

Sin ( A - B ) = Sin A . Cos B - Cos A . Sin B

Given Equation : 

Sin ( A + B ) * Sin ( A - B ) = Sin² A - Sin² B

Proof :

LHS :

Substituting the values from the formula we get,

Sin ( A + B ) * Sin ( A - B )

=> ( Sin A . Cos B + Cos A . Sin B ) * ( Sin A . Cos B  -  Cos A . Sin B )

=>  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 )² - ( Cos A . Sin B )²

=> ( Sin²A . Cos²B ) - ( Cos²A . Sin²B )

=> ( Sin²A ( 1 - Sin²B ) ) - ( ( 1 - Sin²A ) ( Sin²B )

=> Sin²A - Sin²A.Sin²B - ( Sin²B - Sin²A.Sin²B )

=> Sin²A - Sin²A.Sin²B - Sin²B + Sin²A.Sin²B

Sin²A.Sin²B gets cancelled. Then we get,

=> Sin²A - Sin²B

RHS = Sin²A - Sin²B

LHS = RHS

Hence Proved !

Hope my answer helped :-)

Answer 2

$\bold{\huge{\underline{Solution-}}}$

LHS = Sin (A+B) × Sin (A-B) = Sin²A - Sin²B

= 1/2 [2 sin(A + B ) . sin (A - B )]

= 1/2[cos{(A + B ) - (A - B)} - cos{( A + B) + (A - B )}]

= 1/2 [cos(A + B - A + B ) - cos 2A]

= 1/2 ( cos 2B - cos2A)

= 1/2 [( 1- 2sin²B) - (1 - 2sin²A)]

= 1/2( 1- 2sin²B - 1+2sin²A)

= 1/2 (2sin²A - 2sin²B)

= sin²A - sin²B

Alternative method-1

RHS = Sin²A - Sin²B

$\sf \: = \frac{1}{2} (2 {sin}^{2}A \: - 2 {sin}^{2} B) \\ \\ = \sf \: \frac{1}{2} \{ (1 - cos2A) - (1 - cos2B)\} \\ \\ = \sf \: \frac{1}{2} (1 - cos2A - 1 + cos2B) \\ \\ = \sf \: \frac{1}{2} (cos2B - cos2A) \\ \\ = \sf \: \frac{1}{2} \times 2sin \frac{2B + 2A}{2} .sin \frac{2A - 2B}{2} \\ \\ = \sf \: sin(A + B). \: sin(A - B)$

Alternative Method - 2

LHS = sin²A - sin²B

= ( sinA + sinB )(sinA - sinB)

$\bf = 2sin \frac{A + B}{2} \: cos \frac{A - B}{2} \times 2 \: cos \frac{A + B}{2} \: sin \frac{A - B}{2} \\ \\ = \bf \: \left[ 2sin \frac{A + B}{2} \: cos \frac{A + B}{2} \right]\left[2sin \frac{A - B}{2} \: cos \frac{A - B}{2} \right] \\ \\ = \bf \:sin\left[ 2 \times \frac{A + B}{2} \right]\: . \: sin\left[2 \times \frac{A - B}{2} \right] \\ \\ = \bf \: sin(A + B).sin(A - B)$