# sin(x+y)sin(x-y)=cos^2y-cos^2x

Hello!

To prove this trigonometric identity, we need to know the formulas for sine of a sum and sine of a difference. They are as follows:

`sin ( x + y ) = sin ( x ) cos ( y ) + cos ( x ) sin ( y ),`

`sin ( x - y ) = sin ( x ) cos ( y ) - cos ( x ) sin ( y ).`

Note that the second formula may be obtained from the first if we use `-y` instead of `y` and use the fact that `sin ( -y ) = - sin ( y ), cos ( -y ) = cos ( y ).`

Now we can transform the left part `L` of our identity as follows:

`L = sin (x + y ) sin ( x - y ) =` `= ( sin ( x ) cos ( y ) + cos ( x ) sin ( y ) ) ( sin ( x ) cos ( y ) - cos ( x ) sin ( y ) ).`

Use the formula `( a + b ) ( a - b ) = a^2 - b^2` here to move further:

`L = sin^2 ( x ) cos^2 ( y ) - cos^2 ( x ) sin^2 ( y ) =`

`= ( 1 - cos^2 ( x ) ) cos^2 ( y ) - cos^2 ( x ) (1 - cos^2 ( y )) =`

`= cos^2 ( y ) - cos^2 ( x ) cos^2 ( y ) - cos^2 ( x ) + cos^2 ( x ) cos^2 ( y ) =`

`= cos^2 ( y ) - cos^2 ( x ),` which is what we want (the right part of our identity).

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