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/
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