What is another form of    sin(A+B)sin(A-B)/cos^2(A)cos^2(B)

You may convert the product  `sin(A+B)sin(A-B)`  in a sum such that:

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

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

You need to use the formula of cosine of double angle such that:

`cos(2B) -cos(2A) = 2cos^2 B - 1 - 2cos^2 A + 1`

Reducing like terms yields:

`cos(2B)...

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You may convert the product  `sin(A+B)sin(A-B)`  in a sum such that:

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

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

You need to use the formula of cosine of double angle such that:

`cos(2B) -cos(2A) = 2cos^2 B - 1 - 2cos^2 A + 1`

Reducing like terms yields:

`cos(2B) - cos(2A) = 2(cos^2 B - cos^2 A)`

`sin(A+B)sin(A-B) = (1/2)*[2(cos^2 B - cos^2 A)]`

`sin(A+B)sin(A-B) =(cos^2 B - cos^2 A)`

Hence, substituting `(cos^2 B - cos^2 A) for sin(A+B)sin(A-B)

=> (sin(A+B)sin(A-B))/(cos^2(A)cos^2(B)) = (cos^2 B - cos^2 A)/(cos^2(A)cos^2(B)).`

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