How can the cosine rule be derived?

Sorry in advance for the lack of a diagram, which would have made this explanation so much clearer. Also, the published version of this seems to ignore Unicode formatting, so I will add underscores (_) where a subscript was intended and carets (^) where a superscript was intended. I know this is ugly but I don't see a better way.

Consider a triangle with sides A, B and C. The angle between sides B and C we will call a. The line that meets side A at right angles and passes through the corner of B and C we will call Z. The angle between Z and B we will call a__B and the angle between Z and C we will call a__C, hence,

a=a__B+a__C.

By considering the two right angled triangles, the length of side A can be expressed as follows,

A=Bsin(a__B)+Csin(a__C).

Squaring both sides of this equation, we have,

A^²=B^²sin^²(a__B)+C^²sin^²(a__C)+2BCsin(a__B)sin(a__C).

Next, we note that the cosine double angle formula can be applied to give the following,

cos(a)=cos(a__B+a__C)=cos(a__B)cos(a__C)-sin(a__B)sin(a__C).

Subsituting the sines term into the previous equation gives,

A^²=B^²sin^²(a__B)+C^²sin^²(a__C​)-2BCcos(a)+2BCcos(a__B)cos(a__C).

Using the identity,

sin^²(x)=1-cos^²(x),

we have,

A^²=B^²+C^²-2BCcos(a)-B^²cos^²(a__B)-C^²cos^²(a__C​)+2BCcos(a__B)cos(a__C).

The three rightmost terms will factorise to give,

A^²=B^²+C^²-2BCcos(a)-[Bcos(a__B)-Ccos(a__C)]^².

By simple trigonometry, both Bcos(a__B) and Ccos(a__C) are equal Z, hence the rightmost term vanishes and we are left with the cosine rule,

A^²=B^²+C^²-2BCcos(a).