A number theory problem by shivansh mittal
Number Theory
Level
pending
If cosA+ cosB+cosC=0, sinA+sinB+sinC=0 and A+B+C=180', then the value of cos3A+cos3B+cos3C is:
The answer is -3.
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1 solution
Let us take z1=cosA+i sinA,z2=cosB+i sinB, z3=cosC+i sinC →z1+z2+z3=0 →z1^3+z2^3+z3^3=3.z1.z2.z3 =3e^(i(A+B+C)) →z1^3+z2^3+z3^3=e^i3A+e^i3B+e^i3C →3e^(i(A+B+C))=e^i3A+e^i3B+e^i3C Equating real perts on both sides we get: 3cos(A+B+C)=3cos(pi) →-3=cos3A+cos3B+cos3C