cos a + cos b + cos c = sin a +sin b+ sin c =0then value of cos 3a + cos 3b+ cos 3c is..
Meghendra Agrawal , 6 Years ago
Grade
Algebra> cos a + cos b + cos c = sin a +sin b+ sin...
1 Answers
Arun
Last Activity: 6 Years ago
Dear student
we have
cos A + cos B + cos C = 0 ...1
sin A + sin B + sin C = 0 ....2
multiply 2nd by i and add
(cos A + i sin A ) + ( cos B + i sin B) + (cos C + i sin C) = 0
or e^(iA) + e^(iB) + e^(iC)= 0 as e^ix = cos x+ i sin x
and as if (x+y+z) = 0 then x^3+y^3 + z^3 = 3xyz
so e^(i3A) + e^(i3B ) + e^(i3C) = 3 e^i(A+B+C)
so ( cos 3A + i sin 3 A) + (cos 3B + i sin 3B ) + ( cos 3C + i sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)
or (cos 3A + cos 3 B + cos 3C) + i( sin 3A + sin 3B + sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)
equating real and imaginary parts on both sides we get
cos 3A + cos 3 B + cos 3C = 3 cos (A+B+C)
sin 3A + sin 3B + sin 3C = 3 sin (A+B+ C)
cos A + cos B + cos C = 0 ...1
sin A + sin B + sin C = 0 ....2
multiply 2nd by i and add
(cos A + i sin A ) + ( cos B + i sin B) + (cos C + i sin C) = 0
or e^(iA) + e^(iB) + e^(iC)= 0 as e^ix = cos x+ i sin x
and as if (x+y+z) = 0 then x^3+y^3 + z^3 = 3xyz
so e^(i3A) + e^(i3B ) + e^(i3C) = 3 e^i(A+B+C)
so ( cos 3A + i sin 3 A) + (cos 3B + i sin 3B ) + ( cos 3C + i sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)
or (cos 3A + cos 3 B + cos 3C) + i( sin 3A + sin 3B + sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)
equating real and imaginary parts on both sides we get
cos 3A + cos 3 B + cos 3C = 3 cos (A+B+C)
sin 3A + sin 3B + sin 3C = 3 sin (A+B+ C)
Regards
Arun (askIITians forum expert)
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