Sum and difference identity proof

[Joanna]

Sin2Asin2B= sin^2(A+B) - sin^2(A-B)
I would probably try to solve the right side because it's more complicated and use the sum and difference identities but I just do not know what to do with the square root
For instance sin(A+B)= sinAcosB + cosAsinB would I have all the sin and cos squared because the initial equation is squared?? -> sin^2 =sin^2Acos^2B + cos^2Asin^2B ??

 

[mmm4444bot]

Squaring sums or differences does not work that way!

Think:

(1 + 2)^2

That is (1 + 2)*(1 + 2)

You should know by FOILing or by memorizing the famous factor-pattern that this product is not simply 1^2 + 2^2.

It equals 1^2 + 2*(1*2) + 2^2 .

---

Likewise,

sin²(A + B) = sin(A + B)*sin(A + B)

Replace the expression sin(A + B) in both places above with the expression sin(A)*cos(B) + cos(A)*sin(B) and use FOIL.

 

[Joanna]

Ok so sin^2(A+B) - sin^2(A-B)= sin(A+B)*sin(A+B) - sin(A+B)*sin(A-B) <- is this right (I was not sure about the second part of the equation)
And then I expanded everything and used Foil
And I got to 2sinAsinBcosAcosB and I'm stuck :/
I'm not sure how to get from 2sinAsinBcosAcosB to sin2Asin2B ?? Hopefully I made no error

 

[mmm4444bot]

sin^2(A+B) - sin^2(A-B) = sin(A+B)*sin(A+B) - sin(A+B)*sin(A-B)

is this right

Yes, except for the red + above; it should be a minus.



And then I expanded everything and used Foil
And I got to 2sinAsinBcosAcosB

Looks good, except for the factor 2 above; it should be 4.

To continue from 4*sin(A)*sin(B)*cos(A)*cos(B), look at the double-angle identity for sine.

After all, your goal is sin(2A)*sin(2B), so it would be good to know what other form sin(2*angle) can take.