Using our knowledge of this property from before break we took it to the next level. We know how it is derived from the area but today we look at changing sin(a+b) to sin(a-b). This changed the propert and although it may seem confusing at first the process is quite simple. By simply formulating the property with sin(a-b) you would get sina cos-b + sin-b cos a. Althought this may be true the property will not work with -b. We related to even, odd, and neither symmetry to figure out how -b can be changed. Using the sin and cos graphs we can determine that the cos function is even and the sin function is odd ( even means reflected over the y axis and odd is reflected over the y, then the x). This lets us determine that for cos f(-x)=f(x) and for sin f(-x)=-f(x). So by putting that into the property you end up getting sin(a-b)=sina cosb – sinb cosa. Now that we solved that it was time to try solving for cos. The next problem was fcreating the property from cos(a+b). The best way to solve for this is to use the sin version of the property. Since sin(x) and cos(x) are complimentary (telling you that cos(x)=sin(90-(x))) you can make cos(a+b)=sin(90-(a+b)). Now from here you can do they same thing as you did before to create the property. This gives you cos(a+b)=sin(90-a) cosb – sinb cos(90-a). By using supplements again you can find that this comes out to be, cosa cosb – sinb sina. Finally we had to use all of this to solve for cos(a-b). By using the same procces you get almost the same answer except it is + instead of – in the middle. For this property cos(a-b)=cosa cosb + sinb sina.