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

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### 2 Answers

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)).` **

sin(A+B) sin(A-B)

= (sinA cosB + cosA sinB) (sinA cosB - cosA sinB)

= (sinA cosB)^2 - (cosA sinB)^2

= sin²A cos²B - cos²A sin²B

= (1 - cos²A) cos²B - cos²A (1 - cos²B)

= cos²B - cos²A cos²B - cos²A + cos²A cos²B

= cos²B - cos²A (cos²B - cos²A)/(cos^2(A)cos^2(B) = 1/cos^2(A) - 1/cos^2(B)

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