Question 1198384
<font color=black size=3>
Answer: <font color=red>sin^2(A) - sin^2(B)</font>


================================================================================


Explanation:


There are at least two methods we can follow to prove that 
sin(A+B)sin(A-B) = sin^2(A)-sin^2(B)
is an identity


---------------------


Method 1


Let
m = sin(A)cos(B)
n = cos(A)sin(B)


sin(A+B)sin(A-B) = [ sin(A)cos(B)+cos(A)sin(B) ] * [ sin(A)cos(B)-cos(A)sin(B) ]
sin(A+B)sin(A-B) = (m+n)(m-n)
sin(A+B)sin(A-B) = m^2 - n^2
sin(A+B)sin(A-B) = (sin(A)cos(B))^2 - (cos(A)sin(B))^2
sin(A+B)sin(A-B) = sin^2(A)cos^2(B) - cos^2(A)sin^2(B)
sin(A+B)sin(A-B) = sin^2(A)(1-sin^2(B)) - (1-sin^2(A))sin^2(B)
sin(A+B)sin(A-B) = sin^2(A)-sin^2(A)sin^2(B) - sin^2(B)+sin^2(A)sin^2(B)
sin(A+B)sin(A-B) = sin^2(A) - sin^2(B)



The relevant identities that I used were:
sin(A+B) = sin(A)cos(B)+cos(A)sin(B)
sin(A-B) = sin(A)cos(B)-cos(A)sin(B)
cos^2(A) = 1 - sin^2(A) which is from sin^2(A)+cos^2(A) = 1


---------------------


Method 2


Another identity to use is
sin(u)*sin(v) = 0.5 * [ cos(u-v) - cos(u+v) ]
which should be somewhere on your list of trig identities.


Let
u = A+B
v = A-B
So,
sin(u)*sin(v) = 0.5 * [ cos(u-v) - cos(u+v) ]
sin(A+B)*sin(A-B) = 0.5 * [ cos(A+B-(A-B)) - cos(A+B+A-B) ]
sin(A+B)*sin(A-B) = 0.5 * [ cos(2B) - cos(2A) ]
sin(A+B)*sin(A-B) = 0.5 * [ (1-2sin^2(B))-(1-2sin^2(A)) ]
sin(A+B)*sin(A-B) = 0.5 * [ 1-2sin^2(B)-1+2sin^2(A) ]
sin(A+B)*sin(A-B) = 0.5 * [ -2sin^2(B)+2sin^2(A) ]
sin(A+B)*sin(A-B) = 0.5 * [ 2sin^2(A)-2sin^2(B) ]
sin(A+B)*sin(A-B) = sin^2(A)-sin^2(B)


Edit:
Here's a handy list of trig identities
<a href = "https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf">https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf</a>
</font>