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

2 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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)).`

academy633's profile pic

academy633 | College Teacher | (Level 1) Salutatorian

Posted on

 

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)
Sources:

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question