Trigonometry Identity

Date: 02/20/98 at 15:11:53
From: Shweta
Subject: TRIGONOMETRY - Solutions to triangles

Q: In any triangle ABC, prove that 
a*cos(A) + b*cos(B) + c*cos(C) = 4R*sin(A)*sin(B)*sin(C).

I tried using the sine and cosine formulae and
4R = abc/(area of triangle).
I also used 1/2 bc sin(A) = area of triangle.

Please help me out, 
Thank You.

Date: 02/20/98 at 18:31:38
From: Doctor Anthony
Subject: Re: TRIGONOMETRY - Solutions to triangles

We can use the Sine Rule

   a/sin(A) = b/sin(B) = c/sin(C) = 2R

which gives us

   a = 2Rsin(A),

and simililarly for b and c.  Substituting these values for a, b, and 
c into the left side of the equation we want to prove, then 
simplifying and rearranging, we have

   2R[sin(A)cos(A) + sin(B)cos(B) + sin(C)cos(C)] =  
     
      4Rsin(A)*sin(B)*sin(C)

   sin(2A) + sin(2B) + 2sin(C)cos(C) = 4sin(A)sin(B)sin(C)

   2sin(A+B)cos(A-B) + 2sin(C)cos(C) = 4sin(A)sin(B)sin(C)

Now,

   sin(A+B) = sin(180-A-B) = sin(C)

   sin(C)cos(A-B) + sin(C)cos(C) = 2sin(A)sin(B)sin(C)

Divide out by sin(C)

   cos(A-B) + cos(C) = 2sin(A)sin(B)

   2cos((A+C-B)/2)cos((B+C-A)/2) = 2sin(A)sin(B)

Divide by 2, and then since 

                   A+B+C = 180 
                   A+C-B = 180-2B   
               (A+C-B)/2 = 90-B, and similarly  
               (B+C-A)/2 = 90-A,

we have

   cos(90-B)cos(90-A) = sin(A)sin(B)

   sin(B)sin(A) = sin(A)sin(B),  

which is therefore correct. 

-Doctor Anthony, The Math Forum
 http://mathforum.org/dr.math/