Cos(A-B) Theorem Proof

This proof uses the Scalar product to find an expression for Cos (A-B). We use the fact that if a,b are two vectors that
(1) a.b =|a||b| Cos(θ)
let us consider
vector a=ai+bj
vector b=ci+dj
Cos(A-B) theorem proof
The Proof goes as follows p and q are two vectors as shown on the unit circle this means that
|p|=|q|=1 Just write out the Scalar Product of p and q twice.
we can write scalar product of p and q like below
p.q=|p||q|Cos(A-B)=1.1.Cos(A-B)=Cos(A-B)
p.q=CosA CosB + SinA SinB ==> so we can write it as
Cos(A-B)=CosA CosB + SinA SinB


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  Trigonometry