Sin^2(a+b)-sin^2(a-b)=sin(2a)sin(2b )

Step-by-step explanation:

To prove sin(a+b)*sin(a-b)=cos^2b-cos^2a

we simplify the left side  sin(a+b)*sin(a-b) first

sin(a+b) = sin a cos b + cos a sin b

sin(a-b) = sin a cos b - cos a sin b

sin(a+b)*sin(a-b) = (sin a cos b + cos a sin b) x (sin a cos b -cos a sin b)

sin a cos b((sin a cos b + cos a sin b) - cos a sin b (sin a cos b + cos a sin b)

open the bracket

sin a cos b(sin a cos b) + sin a cos b(cos a sin b) -cos a sin b (sin a cos b)+ cos a sin b ( cos a sin b)

sin²a cos²b + sin a cos b cos a sin b - cos a sin b sin a cos b + cos²a sin²b

sin²a cos²b  + 0 + cos²a sin²b

sin²a cos²b + cos²a sin²b

sin²a  = 1-cos² a

sin²b  = 1-cos² b

(1-cos² a)cos² b - cos² a(1-cos² b)

= cos² b - cos² a cos² b - cos² a +cos² a cos² b

choose like terms

cos² b - cos² a - cos² a cos² b + cos² a cos² b = cos² b - cos² a  + 0

cos² b - cos² a

left hand side equals right hand side

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)

10+4√2 is the answer. i dont have the time to explain.