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Cosine Addition Formula


Date: 12/13/97 at 23:32:24
From: Abdul R. Memon
Subject: Important question!

Hello Dr. Math!

My question is how to give a proof for the addition formula for 
cosine. What I mean is, how can you prove it by using right triangles?
Cos(x + y)= cosx cosy - sinx siny - how can you look at two right 
triangles and prove it? Thank you.


Date: 12/14/97 at 03:04:33
From: Doctor Luis
Subject: Re: Important question!

Here goes the answer to your question: I suggest you get hold of some 
pencil and paper right now so you can draw diagrams and follow my 
answer.

Note 1: Here, we shall refer to line segments and to their lengths
interchangeably. The distinction between the two is important since 
one is a geometric object and the other one is a number, but for 
notational purposes they will be represented identically. So, LM could 
either denote the line segment "LM" or the length of the line segment 
"LM". The meaning will be clear from the context.

Note 2: Whenever we refer to a right triangle LMN, the order of
the vertices shall be as follows:

        


Thus, the hypotenuse is the side LN, the right angle is LMN (or NML),
and the two other sides are LM and NM

Make sure you draw good diagrams, that way you won't get confused.

  With that said, let's prove the cosine addition theorem:

  Theorem: cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

  Proof:

  We start by defining a triangle ABC on a plane. Now, define
  another triangle ACD. It's obvious that the hypotenuse AC of
  the first triangle is a side on the other triangle, so, instead
  of drawing them separately, draw triangle ACD on top of ABC
  so that the common side coincides. Remember that they are right
  triangles.

  Now, at the very top should be point D. Drop an altitude
  (i.e. a perpendicular line) from point D to the line segment AB.
  Call the intersection point P. So, now you should have a
  third right triangle APD.  

        


  Essentially, what we have done is define three triangles, where
  an angle in the third triangle can be written as the sum of an
  angle in the first triangle and an angle in the second triangle.
  We will use this fact to establish the cosine addition theorem.

  For future reference, here are some formulas you can get from
  the diagram by simply using the definition of sine and cosine
  (remember these!)

               BC                         CD
    sin CAB = ----             sin DAC = ----
               AC                         AD


               AB                         AC
    cos CAB = ----             cos DAC = ----
               AC                         AD


  So, from the diagram, CAB + DAC = DAP. Call CAB = x, DAC = y.
  Then DAP = x+y. Now, our problem is to find cos DAP. We can
  do that with the help of triangle APD.

              AP
   cos APD = ----
              AD

   From the diagram, you can see that AP = AB - PB.

   We know AB. It's just:     AB = AC cos CAB
   We also know AC:           AC = AD cos DAC
   Combining these last two:  AB = AD cos CAB cos DAC

   Finding PB is a little bit trickier.

   Remember the altitude you dropped from pt D to obtain point P?
   Notice that the line PD intersects line AC at a point. Call
   this intersection point Q.

   Now, this point is part of another triangle QCD. Do you see it?

        


   It turns out that angle QDC is equal to angle CAB (we will use
   this later on). To see why this is true, you should notice that
   angles DQC and AQP are opposite angles (they're formed by two
   intersecting lines AC and PD) and thus they are equal.
   Since angles APQ and QCD are right angles, this means that
   angles QAP and QDC are equal (QAP = QDC), since they are part of
   triangles APQ and QCD, respectively (we already proved
   that two of the angles in these triangles were equal, and so
   the third angles are equal). From the diagram, you can see that
   angle QAP is the same as angle CAB, and so QDC = CAB.

   To find PB, we need an extra point. From pt C, draw a perpendicular
   line segment to line PD. Call the intersection point R.
   You should now see another right triangle DRC. 

        


   Then,

                 RC
      sin RDC = ----     or   RC = CD sin RDC
                 DC

    It should be obvious to you that RC = PB (they're opposite
    sides of a rectangle). Noticing that angle RDC = QDC, we
    can re-write this last equation,

       PB = CD sin QDC

    But QDC = CAB, so

       PB = CD sin CAB

    Finally! We know CD:  CD = AD sin DAC. Therefore,

       PB = AD sin DAC sin CAB

    Now we know the length of line AP,

       AP = AB - PB

       AP = AD cos CAB cos DAC - AD sin DAC sin CAB 

    Then,

                   AP     AD cos CAB cos DAC - AD sin DAC sin CAB
        cos APD = ---- = -----------------------------------------
                   AD                      AD

                = cos CAB cos DAC - sin DAC sin CAB

   If you remember that we defined CAB = x, DAC = y,
   and so DAP = x+y, you can rewrite this last equation
   in the more familiar form,

       cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

                                  Quod Erat Demonstratum
                                 (Latin for "which was to be proven")


   Using this method, you should now be able to prove easily the sine
   angle addition formula,

       sin(x+y) = sin(x)cos(y) + sin(y)cos(x)

  Give it a try to see if you can prove it.

  I hope this helped :)

-Doctors Luis and Sarah,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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