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Math Help - Law of Cosines Derivation

  1. #1
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    Law of Cosines Derivation

    None
    Last edited by Gregorio556; March 25th 2008 at 05:41 AM. Reason: dont need it
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  2. #2
    Member Nacho's Avatar
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    Santiago, Chile
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    Quote Originally Posted by Gregorio556 View Post
    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
    <br />
\begin{gathered}<br />
  \sin (x \pm y) = \sin (x)\cos (x) \pm \sin (y)\cos (y) \hfill \\<br />
  \cos (x \pm y) = \cos (x)\cos (y) \mp \sin (x)\sin (y) \hfill \\<br />
   \hfill \\<br />
  {\text{Example:}} \hfill \\<br />
  {\text{ - find sin(15):}} \hfill \\<br />
  \sin (15) = \sin (45 - 30) \hfill \\ <br />
\end{gathered} <br />
    <br />
\begin{gathered}<br />
  \sin (15) = \sin (45)\cos (30) - \sin (30)\cos (45) \hfill \\<br />
  \sin (15) = \frac{{\sqrt 2 }}<br />
{2} \cdot \frac{{\sqrt 3 }}<br />
{2} - \frac{1}<br />
{2} \cdot \frac{{\sqrt 2 }}<br />
{2} \hfill \\<br />
  \boxed{\sin (15) = \frac{{\sqrt 2 }}<br />
{2}\left( {\frac{{\sqrt 3 }}<br />
{2} - \frac{1}<br />
{2}} \right)} \hfill \\ <br />
\end{gathered} <br />

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

    but for this case especial:

    <br />
\begin{gathered}<br />
  \sin (2x) = \sin (x + x) = 2\sin (x)\cos (x) \hfill \\<br />
  \cos (2x) = \cos (x + x) = \cos ^2 (x) - \sin ^2 (x) \hfill \\ <br />
\end{gathered} <br />

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

    <br />
\cos (x) = \cos \left( {\frac{x}<br />
{2} + \frac{x}<br />
{2}} \right) = \cos ^2 \left( {\frac{x}<br />
{2}} \right) - \sin ^2 \left( {\frac{x}<br />
{2}} \right) = \cos ^2 \left( {\frac{x}<br />
{2}} \right) - \left( {1 - \cos ^2 \left( {\frac{x}<br />
{2}} \right)} \right)<br />
    <br />
\begin{gathered}<br />
  \cos (x) = 2\cos ^2 \left( {\frac{x}<br />
{2}} \right) - 1 \hfill \\<br />
  \cos \left( {\frac{x}<br />
{2}} \right) = \sqrt {\frac{{\cos (x) + 1}}<br />
{2}}  \hfill \\<br />
   \hfill \\<br />
  but: \hfill \\ <br />
\end{gathered} <br />

    <br />
\begin{gathered}<br />
  \cos (x) = \cos ^2 \left( {\frac{x}<br />
{2}} \right) - \sin ^2 \left( {\frac{x}<br />
{2}} \right) = 1 - \sin ^2 \left( {\frac{x}<br />
{2}} \right) - \sin ^2 \left( {\frac{x}<br />
{2}} \right) \hfill \\<br />
  \cos (x) = 1 - 2\sin ^2 \left( {\frac{x}<br />
{2}} \right) \hfill \\ <br />
\end{gathered} <br />
    <br />
\sin \left( {\frac{x}<br />
{2}} \right) = \sqrt {\frac{{1 - \cos (x)}}<br />
{2}} <br />

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

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

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

    to tangent is easily deducible but if you want that i post the formulas, only ask for it
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Nacho View Post
    <br />
\begin{gathered}<br />
  \sin (x \pm y) = \sin (x)\cos (x) \pm \sin (y)\cos (y) \hfill \\<br />
  \cos (x \pm y) = \cos (x)\cos (y) \mp \sin (x)\sin (y) \hfill \\<br />
   \hfill \\<br />
  {\text{Example:}} \hfill \\<br />
  {\text{ - find sin(15):}} \hfill \\<br />
  \sin (15) = \sin (45 - 30) \hfill \\ <br />
\end{gathered} <br />
    <br />
\begin{gathered}<br />
  \sin (15) = \sin (45)\cos (30) - \sin (30)\cos (45) \hfill \\<br />
  \sin (15) = \frac{{\sqrt 2 }}<br />
{2} \cdot \frac{{\sqrt 3 }}<br />
{2} - \frac{1}<br />
{2} \cdot \frac{{\sqrt 2 }}<br />
{2} \hfill \\<br />
  \boxed{\sin (15) = \frac{{\sqrt 2 }}<br />
{2}\left( {\frac{{\sqrt 3 }}<br />
{2} - \frac{1}<br />
{2}} \right)} \hfill \\ <br />
\end{gathered} <br />

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

    but for this case especial:

    <br />
\begin{gathered}<br />
  \sin (2x) = \sin (x + x) = 2\sin (x)\cos (x) \hfill \\<br />
  \cos (2x) = \cos (x + x) = \cos ^2 (x) - \sin ^2 (x) \hfill \\ <br />
\end{gathered} <br />

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

    <br />
\cos (x) = \cos \left( {\frac{x}<br />
{2} + \frac{x}<br />
{2}} \right) = \cos ^2 \left( {\frac{x}<br />
{2}} \right) - \sin ^2 \left( {\frac{x}<br />
{2}} \right) = \cos ^2 \left( {\frac{x}<br />
{2}} \right) - \left( {1 - \cos ^2 \left( {\frac{x}<br />
{2}} \right)} \right)<br />
    <br />
\begin{gathered}<br />
  \cos (x) = 2\cos ^2 \left( {\frac{x}<br />
{2}} \right) - 1 \hfill \\<br />
  \cos \left( {\frac{x}<br />
{2}} \right) = \sqrt {\frac{{\cos (x) + 1}}<br />
{2}}  \hfill \\<br />
   \hfill \\<br />
  but: \hfill \\ <br />
\end{gathered} <br />

    <br />
\begin{gathered}<br />
  \cos (x) = \cos ^2 \left( {\frac{x}<br />
{2}} \right) - \sin ^2 \left( {\frac{x}<br />
{2}} \right) = 1 - \sin ^2 \left( {\frac{x}<br />
{2}} \right) - \sin ^2 \left( {\frac{x}<br />
{2}} \right) \hfill \\<br />
  \cos (x) = 1 - 2\sin ^2 \left( {\frac{x}<br />
{2}} \right) \hfill \\ <br />
\end{gathered} <br />
    <br />
\sin \left( {\frac{x}<br />
{2}} \right) = \sqrt {\frac{{1 - \cos (x)}}<br />
{2}} <br />

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

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

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

    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
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  4. #4
    Newbie
    Joined
    Mar 2008
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    I have found several ways of doing it already.
    Last edited by Gregorio556; March 25th 2008 at 05:42 AM. Reason: none
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