Question

Prove geometrically that $\cos (x+y)=\cos x\cos y-\sin x\sin y$.

Answer 1

Let us take a circle of radius one and let us take $2$ points $P$ and $Q$ such that $P$ is at an angle $x$ and $Q$ at an angle $y$

as shown in the diagram

Therefore, the co-ordinates of $P$ and $Q$ are $P(\cos x,\sin x),Q(\cos y,\sin y)$

Now the distance between $P$ and $Q$ is:

${\left(PQ\right)}^2={(\cos x-\cos y)}^2+{(\sin x-\sin y)}^2=2-2(\cos x.\cos y+\sin x.\sin y)$

Now the distance between $P$ and $Q$ u\sin g \cos ine formula is

${\left(PQ\right)}^2=1^2{+1}^2-2\cos (x-y)=2-2\cos (x-y)$

Comparing both we get

$\cos (x-y)=\cos \left(x\right)\cos \left(y\right)+\sin \left(x\right)\sin \left(y\right)$

Substituting $y$ with $-y$ we get

$\cos (x+y)=\cos x\cos y-\sin x\sin y$