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).
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