sin(A+B)sin(A-B)=?

Answer 1

The formula for sin(a + b) is sin(a)cos(b) + sin(b)cos(a).
The formula for sin(a - b) is sin(a)cos(b) - sin(b)cos(a).

So we have:

sin(A + B)sin(A - B)
=[sin(A)cos(B) + sin(B)cos(A)][sin(A)cos(B) - sin(B)cos(A)]

This equation represents the product of conjugates, so the result should be a difference of squares:

=[sin(A)cos(B)]^2 - [sin(B)cos(A)]^2

=[sin^2(A)cos^2(B) - sin^2(B)cos^2(A)]

Answer 2

sin²acos²b - cos²asin²b doesn't simplify far enough, says Pythagoras:

sin²acos²b - cos²asin²b
= sin²a(1-sin²b) - cos²asin²b
= sin²a - sin²asin²b - cos²asin²b
= sin²a - sin²b(sin²a + cos²a)
= sin²a - sin²b

Answer 3

=sin2A-sin2B=cos2B-cos2A
here 2 is the power or square

Answer 4

the formula for sin(x)siny(y)=
(cos(x-y)-cos(x+y))/2
here x=A+B,y=A-B
x-y=A+B-(A-B)=2B
x+y=A+B+(A-B)=2A
so
sin(A+B)sin(A-B)=
(cos(2B)-cos(2A))/2

Answer 5

sin(A+B)sin(A-B)
=sinAcosB+cosAsinB)(sinAcosB-cosAsinB)
=sin^2Acos^2B-cos^2Asin^2B
=sin^2A(1-sin^2B)-(1-sin^2A)sin^2B
=sin^2Asin^2Asin^2B-sin^2B+sin^2Asin^2B
=sin^2A-sin^2B
You can also prove it further as follows
sin^2A-sin^2B
=(1-c0s^2A)-(1-cos^2B)
=1-cos^2A-1+cos^2B
=cos^2 B-cos^2 A

Answer 6

(sinAcosB + cosAsinB)(sinAcosB - cosAsinB) =

sin^2Acos^2B -cos^2Asin^2B

Answer 7

sin(A+B)sin(A-B)=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)
=>sin^2Acos^2B-cos^2Asin^2B
=>(1-cos^2A)cos^2B - cos^2A(1-cos^2B)
=>cos^2B-cos^2A

Answer 8

½(cos2A-cos2B)

Answer 9

ok...