Question

пж помогите sin²a - sin²B​

Answer 1

Ответ:

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 :-)