Trig identies ..asap ! (1.)2 sin a sin b = a. cos(a-b)-cos(a+b) b. sin(a-b)+tan(a+b) c. sin (a+b)+cos(a-b) ( a - cos a)^2 a. 1-sin2a b. -1+cos a c. sec a +1

(1.)This relies on the angle-sum/difference identity
cos(a+-b)=cos(a)*cos(b)+-sin(a)*sin(b)

2 sin A sin B =
split product into sum:
sin(A)*sin(B)+sin(A)*sin(B)=
add cos(A)*cos(B)-cos(A)*cos(B)=0 -> doesn't break equality
sin(A)*sin(B)+sin(A)*sin(B)+cos(A)*cos(B)-cos(A)*cos(B)=
reorder a bit
cos(A)*cos(B)+sin(A)*sin(B)-cos(A)*cos(B)sin(A)*sin(B)=
add some [] to make the application of the above identity clearer, also note the flipping of +/- on the right one, because of the - in front of the []
[cos(A)*cos(B)+sin(A)*sin(B)] -[ cos(A)*cos(B)-sin(A)*sin(B)]=
A. cos(A-B)-cos(A+B)

and is also listed here (among lots of other identities):
en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities

(2.)This relies on the double angle identity
sin(2x)=2*sin(x)*cos(x)
 and half-angle identities
(1/2)[1-cos(2x)]=sin²(x)
(1/2)[1+cos(2x)]=cos²(x)

(sin A - cos A)^2=
sin²(A)-2*sin(A)*cos(A)+cos²(A)=
apply half angle identities
(1/2)[1-cos(2A)]-2*sin(A)*cos(A)+(1/2)[1+cos(2A)]=
(1/2)[1-cos(2A)+1+cos(2A)]-2*sin(A)*cos(A)=
(1/2)[2]-2*sin(A)*cos(A)=
1-2*sin(A)*cos(A)=
apply double angle identity
1-sin(2A)
A. 1-sin2A

Okay so n is going to represent our first number. since the second number of five times greater than n, the second number will be represented as 5n. for us to figure out the two  numbers we need to have  the value of n first. this is going to be done by adding the two numbers together, n and 5n. it can be simplified as 6n, which the problem states is equal to 72. divide 72 by 6n and you can solve for n.  n+5n=726n=72n=(72/6)n=12  now that we know the value of n, we have to multiply 5 times 12 which equals 60. so our two numbers/integers are 12 and 60