Question

How do you prove that `cos(x-y) = cosxcosy + sinxsiny`?

Answer 1

`cos (a - b) = cos(a)*cos(b) + sin(a)*sin(b)`
can be demonstrated by first showing that
`cos (a + b) = cos(a)*cos(b) - sin(a)*sin(b)`
and then doing the conversion using the CAST principle as indicated.

1. I'm sure there are other ways to do this; but this is what I came up with. (it is pretty long).
2. My apologies for using `a` and `b` instead of `x` and `y`; I drew the diagrams below before checking what variables had been used in the request.

Part 1 : Show `cos (a + b) = cos(a)*cos(b) - sin(a)*sin(b)`

A triangle XQP has been constructed along the hypotenuse of triangle XYQ with angle `a` above angle `b` as in the diagram.

The line segment XP is identified as the unit length for all measurements in this system.

A rectangle is constructed with base XY by extending the line from Y through Q until a point Z is reached where PZ is parallel to the bottom (XY) (completion of the rectangle establishes point W)

Within triangle XQP is is clear that (since `|XP| = 1`)
`|XQ| = cos(a)`
and
`|PQ| = sin(a)`

Therefore, in triangle XYQ
`|XY| = cos(b)*cos(a)` (`cos(b)` scaled up by the `cos(a)`)

Similarly in triangle QZP
`|PZ| = sin(a)*sin(b)`

Since WZ is parallel to XY (by construction)
angle XPW = angle PXY = `a+b`
and
`|WP| = cos(a+b)`

From the diagram
`cos(a+b) + sin(a)*sin(b) = cos(a)*cos(b)`
or
`cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b)`

Part 2 : Show that if `cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b)`
then `cos(a-b) = cos(a)*cos(b) + sin(a)*sin(b)`

`cos(a-b) = cos(a + (-b))`
so we can substitute to get
`cos(a-b) = cos(a)*cos(-b) - sin(a)*sin(-b)`

By the CAST quadrant diagram for trig. signs (below) we can see that
`cos(-b) = cos(b)`
and
`sin(-b) = -sin(b)`

Therefore, we can write:

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

or
`cos (a - b) = cos(a)*cos(b) + sin(a)*sin(b)`