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

Feb 12, 2015

$\cos \left(a - b\right) = \cos \left(a\right) \cdot \cos \left(b\right) + \sin \left(a\right) \cdot \sin \left(b\right)$
can be demonstrated by first showing that
$\cos \left(a + b\right) = \cos \left(a\right) \cdot \cos \left(b\right) - \sin \left(a\right) \cdot \sin \left(b\right)$
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 $\cos \left(a + b\right) = \cos \left(a\right) \cdot \cos \left(b\right) - \sin \left(a\right) \cdot \sin \left(b\right)$

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 $| X P | = 1$)
$| X Q | = \cos \left(a\right)$
and
$| P Q | = \sin \left(a\right)$

Therefore, in triangle XYQ
$| X Y | = \cos \left(b\right) \cdot \cos \left(a\right)$ ($\cos \left(b\right)$ scaled up by the $\cos \left(a\right)$)

Similarly in triangle QZP
$| P Z | = \sin \left(a\right) \cdot \sin \left(b\right)$

Since WZ is parallel to XY (by construction)
angle XPW = angle PXY = $a + b$
and
$| W P | = \cos \left(a + b\right)$

From the diagram
$\cos \left(a + b\right) + \sin \left(a\right) \cdot \sin \left(b\right) = \cos \left(a\right) \cdot \cos \left(b\right)$
or
$\cos \left(a + b\right) = \cos \left(a\right) \cdot \cos \left(b\right) - \sin \left(a\right) \cdot \sin \left(b\right)$

Part 2 : Show that if $\cos \left(a + b\right) = \cos \left(a\right) \cdot \cos \left(b\right) - \sin \left(a\right) \cdot \sin \left(b\right)$
then $\cos \left(a - b\right) = \cos \left(a\right) \cdot \cos \left(b\right) + \sin \left(a\right) \cdot \sin \left(b\right)$

$\cos \left(a - b\right) = \cos \left(a + \left(- b\right)\right)$
so we can substitute to get
$\cos \left(a - b\right) = \cos \left(a\right) \cdot \cos \left(- b\right) - \sin \left(a\right) \cdot \sin \left(- b\right)$

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

Therefore, we can write:

$\cos \left(a - b\right) = \left(\cos \left(a\right) \cdot \cos \left(b\right)\right) - \left(\sin \left(a\right) \cdot \left(- \sin \left(b\right)\right)\right)$

or
$\cos \left(a - b\right) = \cos \left(a\right) \cdot \cos \left(b\right) + \sin \left(a\right) \cdot \sin \left(b\right)$