# Thread: Law of Cosines Derivation

1. ## Law of Cosines Derivation

None

2. Originally Posted by Gregorio556
Hi,

I have a trig problem which requires that I derive the Law of Cosines using "Chapter 10 Material" from my book. This means that I need to use some of the following in the proof.

Double angle formulas
Half angle formulas
Sum or Difference Formulas
Sum or Difference as a Product

I have absolutely no idea how to do it. Please help if you can. I already know the proofs using the distance formula/fundamental trig. All I need is that in some part of the proof one of the above is used.

Thanks
$
\begin{gathered}
\sin (x \pm y) = \sin (x)\cos (x) \pm \sin (y)\cos (y) \hfill \\
\cos (x \pm y) = \cos (x)\cos (y) \mp \sin (x)\sin (y) \hfill \\
\hfill \\
{\text{Example:}} \hfill \\
{\text{ - find sin(15):}} \hfill \\
\sin (15) = \sin (45 - 30) \hfill \\
\end{gathered}
$

$
\begin{gathered}
\sin (15) = \sin (45)\cos (30) - \sin (30)\cos (45) \hfill \\
\sin (15) = \frac{{\sqrt 2 }}
{2} \cdot \frac{{\sqrt 3 }}
{2} - \frac{1}
{2} \cdot \frac{{\sqrt 2 }}
{2} \hfill \\
\boxed{\sin (15) = \frac{{\sqrt 2 }}
{2}\left( {\frac{{\sqrt 3 }}
{2} - \frac{1}
{2}} \right)} \hfill \\
\end{gathered}
$

$
\begin{gathered}
{\text{ - find cos(75)}} \hfill \\
{\text{cos(75)}} = \cos (45 + 30) \hfill \\
\cos (75) = \cos (45)\cos (30) - \sin (45)\cos (30) \hfill \\
etc... \hfill \\
\end{gathered}
$

but for this case especial:

$
\begin{gathered}
\sin (2x) = \sin (x + x) = 2\sin (x)\cos (x) \hfill \\
\cos (2x) = \cos (x + x) = \cos ^2 (x) - \sin ^2 (x) \hfill \\
\end{gathered}
$

that are Half angle formulas and I show you the born of this formulas, the Half angle formulas born too for this

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

$
\begin{gathered}
\cos (x) = 2\cos ^2 \left( {\frac{x}
{2}} \right) - 1 \hfill \\
\cos \left( {\frac{x}
{2}} \right) = \sqrt {\frac{{\cos (x) + 1}}
{2}} \hfill \\
\hfill \\
but: \hfill \\
\end{gathered}
$

$
\begin{gathered}
\cos (x) = \cos ^2 \left( {\frac{x}
{2}} \right) - \sin ^2 \left( {\frac{x}
{2}} \right) = 1 - \sin ^2 \left( {\frac{x}
{2}} \right) - \sin ^2 \left( {\frac{x}
{2}} \right) \hfill \\
\cos (x) = 1 - 2\sin ^2 \left( {\frac{x}
{2}} \right) \hfill \\
\end{gathered}
$

$
\sin \left( {\frac{x}
{2}} \right) = \sqrt {\frac{{1 - \cos (x)}}
{2}}
$

now for Sum or Difference as a Product, is only an identity

$
\begin{gathered}
\sin (x) + \sin (y) = 2\sin \left( {\frac{{x + y}}
{2}} \right)\cos \left( {\frac{{x - y}}
{2}} \right) \hfill \\
\sin (x) - \sin (y) = 2\sin \left( {\frac{{x - y}}
{2}} \right)\cos \left( {\frac{{x + y}}
{2}} \right) \hfill \\
\end{gathered}
$

$
\begin{gathered}
\cos (x) + \cos (y) = 2\cos \left( {\frac{{x + y}}
{2}} \right)\cos \left( {\frac{{x - y}}
{2}} \right) \hfill \\
\cos (x) - \cos (y) = - 2\sin \left( {\frac{{x + y}}
{2}} \right)\sin \left( {\frac{{x - y}}
{2}} \right) \hfill \\
\end{gathered}
$

to tangent is easily deducible but if you want that i post the formulas, only ask for it

3. Originally Posted by Nacho
$
\begin{gathered}
\sin (x \pm y) = \sin (x)\cos (x) \pm \sin (y)\cos (y) \hfill \\
\cos (x \pm y) = \cos (x)\cos (y) \mp \sin (x)\sin (y) \hfill \\
\hfill \\
{\text{Example:}} \hfill \\
{\text{ - find sin(15):}} \hfill \\
\sin (15) = \sin (45 - 30) \hfill \\
\end{gathered}
$

$
\begin{gathered}
\sin (15) = \sin (45)\cos (30) - \sin (30)\cos (45) \hfill \\
\sin (15) = \frac{{\sqrt 2 }}
{2} \cdot \frac{{\sqrt 3 }}
{2} - \frac{1}
{2} \cdot \frac{{\sqrt 2 }}
{2} \hfill \\
\boxed{\sin (15) = \frac{{\sqrt 2 }}
{2}\left( {\frac{{\sqrt 3 }}
{2} - \frac{1}
{2}} \right)} \hfill \\
\end{gathered}
$

$
\begin{gathered}
{\text{ - find cos(75)}} \hfill \\
{\text{cos(75)}} = \cos (45 + 30) \hfill \\
\cos (75) = \cos (45)\cos (30) - \sin (45)\cos (30) \hfill \\
etc... \hfill \\
\end{gathered}
$

but for this case especial:

$
\begin{gathered}
\sin (2x) = \sin (x + x) = 2\sin (x)\cos (x) \hfill \\
\cos (2x) = \cos (x + x) = \cos ^2 (x) - \sin ^2 (x) \hfill \\
\end{gathered}
$

that are Half angle formulas and I show you the born of this formulas, the Half angle formulas born too for this

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

$
\begin{gathered}
\cos (x) = 2\cos ^2 \left( {\frac{x}
{2}} \right) - 1 \hfill \\
\cos \left( {\frac{x}
{2}} \right) = \sqrt {\frac{{\cos (x) + 1}}
{2}} \hfill \\
\hfill \\
but: \hfill \\
\end{gathered}
$

$
\begin{gathered}
\cos (x) = \cos ^2 \left( {\frac{x}
{2}} \right) - \sin ^2 \left( {\frac{x}
{2}} \right) = 1 - \sin ^2 \left( {\frac{x}
{2}} \right) - \sin ^2 \left( {\frac{x}
{2}} \right) \hfill \\
\cos (x) = 1 - 2\sin ^2 \left( {\frac{x}
{2}} \right) \hfill \\
\end{gathered}
$

$
\sin \left( {\frac{x}
{2}} \right) = \sqrt {\frac{{1 - \cos (x)}}
{2}}
$

now for Sum or Difference as a Product, is only an identity

$
\begin{gathered}
\sin (x) + \sin (y) = 2\sin \left( {\frac{{x + y}}
{2}} \right)\cos \left( {\frac{{x - y}}
{2}} \right) \hfill \\
\sin (x) - \sin (y) = 2\sin \left( {\frac{{x - y}}
{2}} \right)\cos \left( {\frac{{x + y}}
{2}} \right) \hfill \\
\end{gathered}
$

$
\begin{gathered}
\cos (x) + \cos (y) = 2\cos \left( {\frac{{x + y}}
{2}} \right)\cos \left( {\frac{{x - y}}
{2}} \right) \hfill \\
\cos (x) - \cos (y) = - 2\sin \left( {\frac{{x + y}}
{2}} \right)\sin \left( {\frac{{x - y}}
{2}} \right) \hfill \\
\end{gathered}
$

to tangent is easily deducible but if you want that i post the formulas, only ask for it
forgive me, i'm a bit tired, but i don't see how what you did helps us to derive the law of cosines.

@Gregorio556: there is no sense in someone re-inventing the wheel. the proof to the cosine rule is something one can easily find by doing a google search or using some other search engine. you may even find several ways of dong it. good luck. you can come back here if you have trouble with one of the derivations and we can work through it

4. I have found several ways of doing it already.