SOLUTION: If cosA+cosB+cosC=0, prove that cos3A+cos3B+cos3C=12cosAcosBcosC

First of all, we need to prove that 

if x + y + z = 0 then  = 3xyz 

x + y + z = 0 

x + y = - z 

 =  

 =  

 = 

Add 3xyz -3xyz (that is 0) to the right hand side 

 =  

Factor -3xy from the first three terms on the right hand side 

 = -3xy(x + y + z) + 3xyz 

 = -3xy(0) + 3xyz 

 = 3xyz 

------- 

Now let x = cosA, y = cosB, z = cosC 

so x + y + z = 0 

Since cos3A =  , then cos3A =  

Similarly, cos3B =  

and cos3C =  

Add all three equations, we have: 

cos3A+cos3B+cos3C =  

=  

= 4(3xyz) - 3(0) 

= 12xyz 

= 12cosAcosBcosC 

Q.E.D 


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