1a shortcut for mcq's
Assume A = 120, B = 240, C = 360
cos A + cos B + cos C = 0
sin A + sin B + sin C = 0
cos 3A + cos 3B + cos 3C = 3
3Cos (A + B + C) = 3
3Sin (A + B + C) = 0
;-)
If sinA+ sin B + sin C = cos A + cos B + cos C = 0
Prove that :
1) sin 3A + sin 3B + sin 3C = 3sin (A+B+C)
2) cos 3A + cos 3B + cos 3C = 3cos (A+B+C)
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7 Answers
106Let Z1 = cosA + isinA
Z2 = cosB + i sinB
Z3 = cosC + isinC
From givn conditions, z1+z2+z3 = 0
and 1/z1 + 1/z2 + 1/z3 = 0
z13+z23+z33 = 3z1z2z3 (as z1+z2+z3=0)
So, putting the relevant expansions we get the result
1
4z1 = cis A
z2 = cis B
z3 = cis C
z1 + z2 + z3 = 0
=> z13 + z23 + z33 = 3z1 z2 z3
=> cos3A + cos3B + cos3C + i(sin3A + sin3B + sin3C) = 3cos(A + B + C) + 3isin(A+B+C)
Equating real n imaginary parts v get the required result
62The one from Uttara is the best..
asish do we really need 1/z+1/z2+1/z3 =0 condition?
106no not at all, but that comes in handy for finding out other results