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Arg of sin(16∏/15)+i(1+ cos(16∏/15)
hi here u can write 1+ cos(16∏/15) = 2cos2 (8∏/15) ......[1+cosx = 2cos^2(x/2).................] also write sin(16∏/15) = 2sin(8∏/15)cos(8∏/15) now d expression becomes 2cos(8∏/15)[sin(8∏/15) +i cos(8∏/15) ] so ur arg wud be simply 8∏/15
hi
here u can write 1+ cos(16∏/15) = 2cos2 (8∏/15) ......[1+cosx = 2cos^2(x/2).................]
also write sin(16∏/15) = 2sin(8∏/15)cos(8∏/15)
now d expression becomes
2cos(8∏/15)[sin(8∏/15) +i cos(8∏/15) ]
so ur arg wud be simply 8∏/15
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