Question

A Micro Tutorial

Answer 1

What is a law of cosines and it's proof Tutorial.

Law of cosines is the law of geometry. It states \[c^2=a^2+b^2-2ab \cos \gamma \], where \[\gamma \] denotes the angle contained between sides of lengths a and b and opposite the side of length c.

\(\huge\text{1st Proof:}\)
Lets look at the triangle ABC. From point C an altitude is drawn.



From a triangle ADC the following things flow:
\[AD=b \cos{a}\]
\[DB=c-b \cos{a}\]
\[h^2 = b^2 – (b* cos~~{a})^2\]
\[h^2 = a^2-(c – b* cos~~{a})^2\]
Equal these equations:
\[ b^2 – (b* cos~~{a})^2=a^2-(c – b* cos~~{a})^2\]
which is equal to
\[a^2=b^2+c^2-2bc~cos~~{a}\]




\(\huge\text{2nd Proof:}\)

Let's draw a triangle on a coordinate grid such way, that A will lie in the origin, AB will lie on some lin OX. AB=c, AC=b, CB=a, angle CAB=\[\alpha \].
Thus the coordinates are A(0, 0), B(c, 0). Using sin and cos and using that AC=b, input the coordinates of C. C(b * cos a, b * sin a).
\[a^2=(b~cos~{a} -c)^2 +b^2*~sin^2 {a}\]
\[a^2=b^2 cos^2~{a} – 2bc~cos~{a} + c^2 + b^2~sin^2~{a}\]
\[a^2=b^2*(cos^2 ~ {a} +sin^2~ {a})+c^2 – 2bc~cos~{a}\]
As well as \[cos^2~{a} + sin^2~{a}=1\] (one of the main trigonometric equations)
=> \[a^2=b^2+c^2-2bc~cos~{a}\]