Question

How do you use the product to sum formulas to write `sin(x+y)cos(x-y)` as a sum or difference?

Answer 1

`sin(x+y)cos(x-y)=1/2(sin2x+sin2y)`

Explanation:

As `cosA=sin(pi/2-A)`

`sin(x+y)cos(x-y)`

= `sin(x+y)sin(pi/2-(x-y))`

= `sin(x+y)sin(pi/2-x+y)`

Now as `2sinAsinB=cos(A-B)-cos(A+B)`

`sin(x+y)sin(pi/2-x+y)`

= `1/2[cos(x+y-(pi/2-x+y))-cos(x+y+(pi/2-x+y))]`

= `1/2[cos(x+y-pi/2+x-y)-cos(x+y+pi/2-x+y)]`

= `1/2[cos(2x-pi/2)-cos(2y+pi/2)]`

= `1/2[cos(pi/2-2x)-cos(pi/2+2y)]`

• as `cos(-A)=cosA`

= `1/2(sin2x+1/2sin2y)`

• as `cos(pi/2-A)=sinA` and `cos(pi/2+A)=-sinA`