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Sine and Cosine Addition Formulas


Date: 11/29/2000 at 00:55:14
From: Phara
Subject: Deriving addition formulas for trig

How do you prove this?

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

What about cos(x+y)? 

If sin(2Q) = a, then what is sin(Q) + cos(Q) in terms of a?


Date: 11/29/2000 at 12:14:08
From: Doctor Rob
Subject: Re: Deriving addition formulas for trig

Thanks for writing to Ask Dr. Math, Phara.

To derive the above identity, draw a diagram as follows. Start with a 
Cartesian rectangular coordinate system with origin O. Pick a point A 
on the positive x-axis. Construct ray OB so that <AOB has measure x. 
Construct ray OC so that <BOC has measure y. Then <AOC has measure 
x + y. Erect a perpendicular to OA at A intersecting OB at D. Erect a 
perpendicular to OB at D intersecting OC at E. Drop a perpendicular 
from E to OA, intersecting it at F, and intersecting OD at G. Now you 
have several right triangles, OAD, ODE, OFE, OFG, and GDE, each 
containing one of the angles with measure x, y, and x + y:

       ^             /C
       |           E/
       |           /\
       |          /|x\
       |         / |  \
       |        /  |   \
       |       /   |    \             B
       |      /    |     \       _,-'
       |     /     |      \D _,-'
       |    /      |     _,o'  |
       |   /      G| _,-'  |
       |  /      _,+'      |
       | /y  _,-'  |       |
       |/_,-'x     |       |
     --o-----------o-------o--->
      O|           F       A

Now let OE have length r. Then, using triangle ODE and the definitions 
of sine and cosine,

     OD = r*cos(y)
     DE = r*sin(y)

Next, using triangle GDE,

     EG = r*sin(y)/cos(x)
     DG = r*sin(y)*tan(x)

That implies that

     OG = OD - DG
        = r*cos(y) - r*sin(y)*tan(x)
        = r*[cos(y)-sin(y)*tan(x)]

Now using triangle OFG,

     FG = OG*sin(x)
        = r*[sin(x)*cos(y)-sin(y)*tan(x)*sin(x)]

That tells you the length

     EF = EG + FG
        = r*sin(y)/cos(x) + r*[sin(x)*cos(y)-sin(y)*tan(x)*sin(x)]
        = r*[sin(x)*cos(y)-sin(y)*sin^2(x)/cos(x)+sin(y)/cos(x)]
        = r*[sin(x)*cos(y)+sin(y)*[1-sin^2(x)]/cos(x)]
        = r*[sin(x)*cos(y)+sin(y)*cos^2(x)/cos(x)]
        = r*[sin(x)*cos(y)+cos(x)*sin(y)]

On the other hand, using triangle OFE,

     EF = r*sin(x+y)

Setting these last two equal and dividing out the r gives

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

To derive the cosine identity, consider

     cos(x+y) = sin(pi/2-[x+y])
              = sin([pi/2-x]+[-y])

Now use the sine identity proved above:

     cos(x+y) = sin(Pi/2-x)*cos(-y) + cos(Pi/2-x)*sin(-y)
              = cos(x)*cos(y) - sin(x)*sin(y)

For your other question,

                                  sin(2*Q) = a
                           2*sin(Q)*cos(Q) = a
                       1 + 2*sin(Q)*cos(Q) = 1 + a
     sin^2(Q) + 2*sin(Q)*cos(Q) + cos^2(Q) = 1 + a
                         [sin(Q)+cos(Q)]^2 = 1 + a
                           sin(Q) + cos(Q) = +-sqrt(1+a)

The sign to use is determined by which quadrant Q + pi/4 might lie in. 
If Q + pi/4 is in the first or second quadrant, then the positive sign 
is correct, and otherwise the negative sign is the one to use. This is 
because

     sin(Q) + cos(Q) = sqrt(2)*sin(Q+Pi/4)

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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